MAYBE We are left with following problem, upon which TcT provides the certificate MAYBE. Strict Trs: { sel(s(X), cons(Y, Z)) -> sel(X, Z) , sel(0(), cons(X, Z)) -> X , first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) , first(0(), Z) -> nil() , from(X) -> cons(X, from(s(X))) , sel1(s(X), cons(Y, Z)) -> sel1(X, Z) , sel1(0(), cons(X, Z)) -> quote(X) , quote(sel(X, Z)) -> sel1(X, Z) , quote(s(X)) -> s1(quote(X)) , quote(0()) -> 01() , first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z)) , first1(0(), Z) -> nil1() , quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z)) , quote1(first(X, Z)) -> first1(X, Z) , quote1(nil()) -> nil1() , unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 60 seconds)' failed due to the following reason: We add the following innermost weak dependency pairs: Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2() , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , first^#(0(), Z) -> c_4() , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , first1^#(0(), Z) -> c_12() , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2() , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , first^#(0(), Z) -> c_4() , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , first1^#(0(), Z) -> c_12() , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } Strict Trs: { sel(s(X), cons(Y, Z)) -> sel(X, Z) , sel(0(), cons(X, Z)) -> X , first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z)) , first(0(), Z) -> nil() , from(X) -> cons(X, from(s(X))) , sel1(s(X), cons(Y, Z)) -> sel1(X, Z) , sel1(0(), cons(X, Z)) -> quote(X) , quote(sel(X, Z)) -> sel1(X, Z) , quote(s(X)) -> s1(quote(X)) , quote(0()) -> 01() , first1(s(X), cons(Y, Z)) -> cons1(quote(Y), first1(X, Z)) , first1(0(), Z) -> nil1() , quote1(cons(X, Z)) -> cons1(quote(X), quote1(Z)) , quote1(first(X, Z)) -> first1(X, Z) , quote1(nil()) -> nil1() , unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE We replace rewrite rules by usable rules: Strict Usable Rules: { unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2() , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , first^#(0(), Z) -> c_4() , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , first1^#(0(), Z) -> c_12() , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote^#(s1(X)) -> c_17(unquote^#(X)) , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } Strict Trs: { unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(s) = {1}, Uargs(fcons) = {1, 2}, Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(c_9) = {1}, Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2}, Uargs(c_14) = {1}, Uargs(c_17) = {1}, Uargs(c_19) = {1}, Uargs(fcons^#) = {1, 2} TcT has computed the following constructor-restricted matrix interpretation. [sel](x1, x2) = [2 2] x1 + [2 2] x2 + [0] [1 1] [1 1] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [cons](x1, x2) = [1 0] x1 + [1 0] x2 + [0] [0 0] [0 0] [0] [0] = [0] [0] [first](x1, x2) = [2 2] x1 + [2 1] x2 + [0] [1 1] [1 2] [0] [nil] = [0] [0] [nil1] = [0] [0] [cons1](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [2] [01] = [0] [0] [s1](x1) = [1 0] x1 + [2] [0 0] [0] [unquote](x1) = [2 0] x1 + [1] [0 0] [0] [unquote1](x1) = [2 0] x1 + [1] [0 2] [0] [fcons](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [0] [sel^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 2] [2 2] [2] [c_1](x1) = [1 0] x1 + [1] [0 1] [1] [c_2] = [2] [1] [first^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 1] [2 2] [2] [c_3](x1) = [1 0] x1 + [1] [0 1] [1] [c_4] = [1] [1] [from^#](x1) = [1 2] x1 + [2] [2 1] [2] [c_5](x1) = [1 0] x1 + [2] [0 1] [2] [sel1^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 2] [1 2] [2] [c_6](x1) = [1 0] x1 + [1] [0 1] [1] [c_7](x1) = [1 0] x1 + [2] [0 1] [2] [quote^#](x1) = [0] [0] [c_8](x1) = [1 0] x1 + [1] [0 1] [1] [c_9](x1) = [1 0] x1 + [2] [0 1] [2] [c_10] = [1] [1] [first1^#](x1, x2) = [0 0] x1 + [0 0] x2 + [2] [1 1] [2 2] [2] [c_11](x1, x2) = [1 0] x1 + [1 0] x2 + [1] [0 1] [0 1] [1] [c_12] = [1] [1] [quote1^#](x1) = [0] [0] [c_13](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 1] [0 1] [2] [c_14](x1) = [1 0] x1 + [1] [0 1] [1] [c_15] = [1] [1] [unquote^#](x1) = [2 0] x1 + [2] [1 1] [2] [c_16] = [1] [1] [c_17](x1) = [1 0] x1 + [1] [0 1] [1] [unquote1^#](x1) = [2 1] x1 + [1] [1 1] [2] [c_18] = [0] [1] [c_19](x1) = [1 0] x1 + [2] [0 1] [2] [fcons^#](x1, x2) = [1 0] x1 + [1 0] x2 + [2] [0 0] [0 0] [2] [c_20] = [1] [1] The following symbols are considered usable {unquote, unquote1, fcons, sel^#, first^#, from^#, sel1^#, quote^#, first1^#, quote1^#, unquote^#, unquote1^#, fcons^#} The order satisfies the following ordering constraints: [unquote(01())] = [1] [0] > [0] [0] = [0()] [unquote(s1(X))] = [2 0] X + [5] [0 0] [0] > [2 0] X + [1] [0 0] [0] = [s(unquote(X))] [unquote1(nil1())] = [1] [0] > [0] [0] = [nil()] [unquote1(cons1(X, Z))] = [2 0] X + [2 0] Z + [5] [0 0] [0 0] [4] > [2 0] X + [2 0] Z + [4] [0 0] [0 0] [0] = [fcons(unquote(X), unquote1(Z))] [fcons(X, Z)] = [1 0] X + [1 0] Z + [2] [0 0] [0 0] [0] > [1 0] X + [1 0] Z + [0] [0 0] [0 0] [0] = [cons(X, Z)] [sel^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2] [1 0] [2 0] [2 0] [2] ? [0 0] X + [0 0] Z + [3] [1 2] [2 2] [3] = [c_1(sel^#(X, Z))] [sel^#(0(), cons(X, Z))] = [0 0] X + [0 0] Z + [2] [2 0] [2 0] [2] >= [2] [1] = [c_2()] [first^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2] [1 0] [2 0] [2 0] [2] ? [0 0] X + [0 0] Z + [3] [1 1] [2 2] [3] = [c_3(first^#(X, Z))] [first^#(0(), Z)] = [0 0] Z + [2] [2 2] [2] > [1] [1] = [c_4()] [from^#(X)] = [1 2] X + [2] [2 1] [2] ? [1 0] X + [4] [2 0] [4] = [c_5(from^#(s(X)))] [sel1^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2] [1 0] [1 0] [1 0] [2] ? [0 0] X + [0 0] Z + [3] [1 2] [1 2] [3] = [c_6(sel1^#(X, Z))] [sel1^#(0(), cons(X, Z))] = [0 0] X + [0 0] Z + [2] [1 0] [1 0] [2] >= [2] [2] = [c_7(quote^#(X))] [quote^#(sel(X, Z))] = [0] [0] ? [0 0] X + [0 0] Z + [3] [1 2] [1 2] [3] = [c_8(sel1^#(X, Z))] [quote^#(s(X))] = [0] [0] ? [2] [2] = [c_9(quote^#(X))] [quote^#(0())] = [0] [0] ? [1] [1] = [c_10()] [first1^#(s(X), cons(Y, Z))] = [0 0] X + [0 0] Y + [0 0] Z + [2] [1 0] [2 0] [2 0] [2] ? [0 0] X + [0 0] Z + [3] [1 1] [2 2] [3] = [c_11(quote^#(Y), first1^#(X, Z))] [first1^#(0(), Z)] = [0 0] Z + [2] [2 2] [2] > [1] [1] = [c_12()] [quote1^#(cons(X, Z))] = [0] [0] ? [2] [2] = [c_13(quote^#(X), quote1^#(Z))] [quote1^#(first(X, Z))] = [0] [0] ? [0 0] X + [0 0] Z + [3] [1 1] [2 2] [3] = [c_14(first1^#(X, Z))] [quote1^#(nil())] = [0] [0] ? [1] [1] = [c_15()] [unquote^#(01())] = [2] [2] > [1] [1] = [c_16()] [unquote^#(s1(X))] = [2 0] X + [6] [1 0] [4] ? [2 0] X + [3] [1 1] [3] = [c_17(unquote^#(X))] [unquote1^#(nil1())] = [1] [2] > [0] [1] = [c_18()] [unquote1^#(cons1(X, Z))] = [2 0] X + [2 0] Z + [7] [1 0] [1 0] [6] > [2 0] X + [2 0] Z + [6] [0 0] [0 0] [4] = [c_19(fcons^#(unquote(X), unquote1(Z)))] [fcons^#(X, Z)] = [1 0] X + [1 0] Z + [2] [0 0] [0 0] [2] > [1] [1] = [c_20()] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , sel^#(0(), cons(X, Z)) -> c_2() , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , quote^#(0()) -> c_10() , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , quote1^#(nil()) -> c_15() , unquote^#(s1(X)) -> c_17(unquote^#(X)) } Weak DPs: { first^#(0(), Z) -> c_4() , first1^#(0(), Z) -> c_12() , unquote^#(01()) -> c_16() , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } Weak Trs: { unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE We estimate the number of application of {2,9,13} by applications of Pre({2,9,13}) = {1,6,8,10,11}. Here rules are labeled as follows: DPs: { 1: sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , 2: sel^#(0(), cons(X, Z)) -> c_2() , 3: first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , 4: from^#(X) -> c_5(from^#(s(X))) , 5: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , 6: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , 7: quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , 8: quote^#(s(X)) -> c_9(quote^#(X)) , 9: quote^#(0()) -> c_10() , 10: first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , 11: quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , 12: quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , 13: quote1^#(nil()) -> c_15() , 14: unquote^#(s1(X)) -> c_17(unquote^#(X)) , 15: first^#(0(), Z) -> c_4() , 16: first1^#(0(), Z) -> c_12() , 17: unquote^#(01()) -> c_16() , 18: unquote1^#(nil1()) -> c_18() , 19: unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , 20: fcons^#(X, Z) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , unquote^#(s1(X)) -> c_17(unquote^#(X)) } Weak DPs: { sel^#(0(), cons(X, Z)) -> c_2() , first^#(0(), Z) -> c_4() , quote^#(0()) -> c_10() , first1^#(0(), Z) -> c_12() , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } Weak Trs: { unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { sel^#(0(), cons(X, Z)) -> c_2() , first^#(0(), Z) -> c_4() , quote^#(0()) -> c_10() , first1^#(0(), Z) -> c_12() , quote1^#(nil()) -> c_15() , unquote^#(01()) -> c_16() , unquote1^#(nil1()) -> c_18() , unquote1^#(cons1(X, Z)) -> c_19(fcons^#(unquote(X), unquote1(Z))) , fcons^#(X, Z) -> c_20() } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , unquote^#(s1(X)) -> c_17(unquote^#(X)) } Weak Trs: { unquote(01()) -> 0() , unquote(s1(X)) -> s(unquote(X)) , unquote1(nil1()) -> nil() , unquote1(cons1(X, Z)) -> fcons(unquote(X), unquote1(Z)) , fcons(X, Z) -> cons(X, Z) } Obligation: innermost runtime complexity Answer: MAYBE No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , unquote^#(s1(X)) -> c_17(unquote^#(X)) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'WithProblem' failed due to the following reason: We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 4: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , 5: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , 7: quote^#(s(X)) -> c_9(quote^#(X)) , 8: first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , 11: unquote^#(s1(X)) -> c_17(unquote^#(X)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(c_9) = {1}, Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2}, Uargs(c_14) = {1}, Uargs(c_17) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [sel](x1, x2) = [4] x1 + [4] x2 + [0] [s](x1) = [1] x1 + [3] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [0] = [3] [first](x1, x2) = [4] x1 + [1] x2 + [0] [nil] = [0] [from](x1) = [7] x1 + [0] [sel1](x1, x2) = [7] x1 + [7] x2 + [0] [quote](x1) = [7] x1 + [0] [first1](x1, x2) = [7] x1 + [7] x2 + [0] [nil1] = [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [01] = [0] [quote1](x1) = [7] x1 + [0] [s1](x1) = [1] x1 + [4] [unquote](x1) = [7] x1 + [0] [unquote1](x1) = [7] x1 + [0] [fcons](x1, x2) = [7] x1 + [7] x2 + [0] [sel^#](x1, x2) = [0] [c_1](x1) = [1] x1 + [0] [c_2] = [0] [first^#](x1, x2) = [0] [c_3](x1) = [4] x1 + [0] [c_4] = [0] [from^#](x1) = [0] [c_5](x1) = [1] x1 + [0] [sel1^#](x1, x2) = [3] x1 + [4] x2 + [0] [c_6](x1) = [1] x1 + [7] [c_7](x1) = [2] x1 + [3] [quote^#](x1) = [2] x1 + [0] [c_8](x1) = [2] x1 + [0] [c_9](x1) = [1] x1 + [3] [c_10] = [0] [first1^#](x1, x2) = [3] x1 + [2] x2 + [0] [c_11](x1, x2) = [1] x1 + [1] x2 + [7] [c_12] = [0] [quote1^#](x1) = [2] x1 + [0] [c_13](x1, x2) = [1] x1 + [1] x2 + [0] [c_14](x1) = [1] x1 + [0] [c_15] = [0] [unquote^#](x1) = [2] x1 + [0] [c_16] = [0] [c_17](x1) = [1] x1 + [5] [unquote1^#](x1) = [7] x1 + [0] [c_18] = [0] [c_19](x1) = [7] x1 + [0] [fcons^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_20] = [0] The following symbols are considered usable {sel^#, first^#, from^#, sel1^#, quote^#, first1^#, quote1^#, unquote^#} The order satisfies the following ordering constraints: [sel^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_1(sel^#(X, Z))] [first^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_3(first^#(X, Z))] [from^#(X)] = [0] >= [0] = [c_5(from^#(s(X)))] [sel1^#(s(X), cons(Y, Z))] = [3] X + [4] Y + [4] Z + [9] > [3] X + [4] Z + [7] = [c_6(sel1^#(X, Z))] [sel1^#(0(), cons(X, Z))] = [4] X + [4] Z + [9] > [4] X + [3] = [c_7(quote^#(X))] [quote^#(sel(X, Z))] = [8] X + [8] Z + [0] >= [6] X + [8] Z + [0] = [c_8(sel1^#(X, Z))] [quote^#(s(X))] = [2] X + [6] > [2] X + [3] = [c_9(quote^#(X))] [first1^#(s(X), cons(Y, Z))] = [3] X + [2] Y + [2] Z + [9] > [3] X + [2] Y + [2] Z + [7] = [c_11(quote^#(Y), first1^#(X, Z))] [quote1^#(cons(X, Z))] = [2] X + [2] Z + [0] >= [2] X + [2] Z + [0] = [c_13(quote^#(X), quote1^#(Z))] [quote1^#(first(X, Z))] = [8] X + [2] Z + [0] >= [3] X + [2] Z + [0] = [c_14(first1^#(X, Z))] [unquote^#(s1(X))] = [2] X + [8] > [2] X + [5] = [c_17(unquote^#(X))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) } Weak DPs: { sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , unquote^#(s1(X)) -> c_17(unquote^#(X)) } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { unquote^#(s1(X)) -> c_17(unquote^#(X)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) } Weak DPs: { sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) } Obligation: innermost runtime complexity Answer: MAYBE We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 4: quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , 5: quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , 6: quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) , 7: sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , 8: sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , 9: quote^#(s(X)) -> c_9(quote^#(X)) , 10: first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1}, Uargs(c_6) = {1}, Uargs(c_7) = {1}, Uargs(c_8) = {1}, Uargs(c_9) = {1}, Uargs(c_11) = {1, 2}, Uargs(c_13) = {1, 2}, Uargs(c_14) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [sel](x1, x2) = [4] x2 + [4] [s](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x1 + [1] x2 + [2] [0] = [7] [first](x1, x2) = [2] x1 + [2] x2 + [2] [nil] = [0] [from](x1) = [7] x1 + [0] [sel1](x1, x2) = [7] x1 + [7] x2 + [0] [quote](x1) = [7] x1 + [0] [first1](x1, x2) = [7] x1 + [7] x2 + [0] [nil1] = [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [01] = [0] [quote1](x1) = [7] x1 + [0] [s1](x1) = [1] x1 + [0] [unquote](x1) = [7] x1 + [0] [unquote1](x1) = [7] x1 + [0] [fcons](x1, x2) = [7] x1 + [7] x2 + [0] [sel^#](x1, x2) = [0] [c_1](x1) = [4] x1 + [0] [c_2] = [0] [first^#](x1, x2) = [0] [c_3](x1) = [4] x1 + [0] [c_4] = [0] [from^#](x1) = [0] [c_5](x1) = [1] x1 + [0] [sel1^#](x1, x2) = [4] x2 + [0] [c_6](x1) = [1] x1 + [1] [c_7](x1) = [2] x1 + [1] [quote^#](x1) = [2] x1 + [0] [c_8](x1) = [1] x1 + [7] [c_9](x1) = [1] x1 + [1] [c_10] = [0] [first1^#](x1, x2) = [4] x2 + [0] [c_11](x1, x2) = [1] x1 + [1] x2 + [1] [c_12] = [0] [quote1^#](x1) = [4] x1 + [0] [c_13](x1, x2) = [1] x1 + [1] x2 + [1] [c_14](x1) = [1] x1 + [1] [c_15] = [0] [unquote^#](x1) = [7] x1 + [0] [c_16] = [0] [c_17](x1) = [7] x1 + [0] [unquote1^#](x1) = [7] x1 + [0] [c_18] = [0] [c_19](x1) = [7] x1 + [0] [fcons^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_20] = [0] The following symbols are considered usable {sel^#, first^#, from^#, sel1^#, quote^#, first1^#, quote1^#} The order satisfies the following ordering constraints: [sel^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_1(sel^#(X, Z))] [first^#(s(X), cons(Y, Z))] = [0] >= [0] = [c_3(first^#(X, Z))] [from^#(X)] = [0] >= [0] = [c_5(from^#(s(X)))] [sel1^#(s(X), cons(Y, Z))] = [4] Y + [4] Z + [8] > [4] Z + [1] = [c_6(sel1^#(X, Z))] [sel1^#(0(), cons(X, Z))] = [4] X + [4] Z + [8] > [4] X + [1] = [c_7(quote^#(X))] [quote^#(sel(X, Z))] = [8] Z + [8] > [4] Z + [7] = [c_8(sel1^#(X, Z))] [quote^#(s(X))] = [2] X + [8] > [2] X + [1] = [c_9(quote^#(X))] [first1^#(s(X), cons(Y, Z))] = [4] Y + [4] Z + [8] > [2] Y + [4] Z + [1] = [c_11(quote^#(Y), first1^#(X, Z))] [quote1^#(cons(X, Z))] = [4] X + [4] Z + [8] > [2] X + [4] Z + [1] = [c_13(quote^#(X), quote1^#(Z))] [quote1^#(first(X, Z))] = [8] X + [8] Z + [8] > [4] Z + [1] = [c_14(first1^#(X, Z))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) } Weak DPs: { sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { sel1^#(s(X), cons(Y, Z)) -> c_6(sel1^#(X, Z)) , sel1^#(0(), cons(X, Z)) -> c_7(quote^#(X)) , quote^#(sel(X, Z)) -> c_8(sel1^#(X, Z)) , quote^#(s(X)) -> c_9(quote^#(X)) , first1^#(s(X), cons(Y, Z)) -> c_11(quote^#(Y), first1^#(X, Z)) , quote1^#(cons(X, Z)) -> c_13(quote^#(X), quote1^#(Z)) , quote1^#(first(X, Z)) -> c_14(first1^#(X, Z)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) , from^#(X) -> c_5(from^#(s(X))) } Obligation: innermost runtime complexity Answer: MAYBE We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. DPs: { 1: sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , 2: first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) } Sub-proof: ---------- The following argument positions are usable: Uargs(c_1) = {1}, Uargs(c_3) = {1}, Uargs(c_5) = {1} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [sel](x1, x2) = [7] x1 + [7] x2 + [0] [s](x1) = [1] x1 + [2] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [0] = [0] [first](x1, x2) = [7] x1 + [7] x2 + [0] [nil] = [0] [from](x1) = [7] x1 + [0] [sel1](x1, x2) = [7] x1 + [7] x2 + [0] [quote](x1) = [7] x1 + [0] [first1](x1, x2) = [7] x1 + [7] x2 + [0] [nil1] = [0] [cons1](x1, x2) = [1] x1 + [1] x2 + [0] [01] = [0] [quote1](x1) = [7] x1 + [0] [s1](x1) = [1] x1 + [0] [unquote](x1) = [7] x1 + [0] [unquote1](x1) = [7] x1 + [0] [fcons](x1, x2) = [7] x1 + [7] x2 + [0] [sel^#](x1, x2) = [4] x1 + [0] [c_1](x1) = [1] x1 + [3] [c_2] = [0] [first^#](x1, x2) = [4] x1 + [0] [c_3](x1) = [1] x1 + [5] [c_4] = [0] [from^#](x1) = [0] [c_5](x1) = [4] x1 + [0] [sel1^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_6](x1) = [7] x1 + [0] [c_7](x1) = [7] x1 + [0] [quote^#](x1) = [7] x1 + [0] [c_8](x1) = [7] x1 + [0] [c_9](x1) = [7] x1 + [0] [c_10] = [0] [first1^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_11](x1, x2) = [7] x1 + [7] x2 + [0] [c_12] = [0] [quote1^#](x1) = [7] x1 + [0] [c_13](x1, x2) = [7] x1 + [7] x2 + [0] [c_14](x1) = [7] x1 + [0] [c_15] = [0] [unquote^#](x1) = [7] x1 + [0] [c_16] = [0] [c_17](x1) = [7] x1 + [0] [unquote1^#](x1) = [7] x1 + [0] [c_18] = [0] [c_19](x1) = [7] x1 + [0] [fcons^#](x1, x2) = [7] x1 + [7] x2 + [0] [c_20] = [0] The following symbols are considered usable {sel^#, first^#, from^#} The order satisfies the following ordering constraints: [sel^#(s(X), cons(Y, Z))] = [4] X + [8] > [4] X + [3] = [c_1(sel^#(X, Z))] [first^#(s(X), cons(Y, Z))] = [4] X + [8] > [4] X + [5] = [c_3(first^#(X, Z))] [from^#(X)] = [0] >= [0] = [c_5(from^#(s(X)))] The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { from^#(X) -> c_5(from^#(s(X))) } Weak DPs: { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) } Obligation: innermost runtime complexity Answer: MAYBE The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { sel^#(s(X), cons(Y, Z)) -> c_1(sel^#(X, Z)) , first^#(s(X), cons(Y, Z)) -> c_3(first^#(X, Z)) } We are left with following problem, upon which TcT provides the certificate MAYBE. Strict DPs: { from^#(X) -> c_5(from^#(s(X))) } Obligation: innermost runtime complexity Answer: MAYBE None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Fastest' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'empty' failed due to the following reason: Empty strict component of the problem is NOT empty. 2) 'Polynomial Path Order (PS)' failed due to the following reason: The input cannot be shown compatible 2) 'Fastest (timeout of 5 seconds)' failed due to the following reason: Computation stopped due to timeout after 5.0 seconds. 3) 'Polynomial Path Order (PS)' failed due to the following reason: The input cannot be shown compatible 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'WithProblem (timeout of 30 seconds) (timeout of 60 seconds)' failed due to the following reason: Computation stopped due to timeout after 30.0 seconds. 2) 'Best' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Polynomial Path Order (PS) (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 2) 'bsearch-popstar (timeout of 60 seconds)' failed due to the following reason: The input cannot be shown compatible 3) 'Fastest (timeout of 5 seconds) (timeout of 60 seconds)' failed due to the following reason: None of the processors succeeded. Details of failed attempt(s): ----------------------------- 1) 'Bounds with minimal-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. 2) 'Bounds with perSymbol-enrichment and initial automaton 'match'' failed due to the following reason: match-boundness of the problem could not be verified. Arrrr..