YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , terms(X) -> n__terms(X) , sqr(0()) -> 0() , sqr(s(X)) -> s(add(sqr(X), dbl(X))) , s(X) -> n__s(X) , add(0(), X) -> X , add(s(X), Y) -> s(add(X, Y)) , dbl(0()) -> 0() , dbl(s(X)) -> s(s(dbl(X))) , first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(s(0())) -> 0() , half(s(s(X))) -> s(half(X)) , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) Arguments of following rules are not normal-forms: { sqr(s(X)) -> s(add(sqr(X), dbl(X))) , add(s(X), Y) -> s(add(X, Y)) , dbl(s(X)) -> s(s(dbl(X))) , first(s(X), cons(Y, Z)) -> cons(Y, n__first(X, activate(Z))) , half(s(0())) -> 0() , half(s(s(X))) -> s(half(X)) } All above mentioned rules can be savely removed. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , terms(X) -> n__terms(X) , sqr(0()) -> 0() , s(X) -> n__s(X) , add(0(), X) -> X , dbl(0()) -> 0() , first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [terms](x1) = [1] x1 + [5] [cons](x1, x2) = [1] x1 + [1] x2 + [0] [recip](x1) = [1] x1 + [0] [sqr](x1) = [1] [n__terms](x1) = [1] x1 + [7] [n__s](x1) = [1] x1 + [7] [0] = [7] [s](x1) = [1] x1 + [7] [add](x1, x2) = [1] x1 + [1] x2 + [7] [dbl](x1) = [1] x1 + [7] [first](x1, x2) = [1] x1 + [1] x2 + [3] [nil] = [1] [n__first](x1, x2) = [1] x1 + [1] x2 + [6] [activate](x1) = [1] x1 + [6] [half](x1) = [1] x1 + [7] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [5] ? [1] N + [15] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [5] ? [1] X + [7] = [n__terms(X)] [sqr(0())] = [1] ? [7] = [0()] [s(X)] = [1] X + [7] >= [1] X + [7] = [n__s(X)] [add(0(), X)] = [1] X + [14] > [1] X + [0] = [X] [dbl(0())] = [14] > [7] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [3] ? [1] X1 + [1] X2 + [6] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [10] > [1] = [nil()] [activate(X)] = [1] X + [6] > [1] X + [0] = [X] [activate(n__terms(X))] = [1] X + [13] > [1] X + [11] = [terms(activate(X))] [activate(n__s(X))] = [1] X + [13] >= [1] X + [13] = [s(activate(X))] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [12] ? [1] X1 + [1] X2 + [15] = [first(activate(X1), activate(X2))] [half(0())] = [14] > [7] = [0()] [half(dbl(X))] = [1] X + [14] > [1] X + [0] = [X] 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 YES(O(1),O(n^1)). Strict Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , terms(X) -> n__terms(X) , sqr(0()) -> 0() , s(X) -> n__s(X) , first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) } Weak Trs: { add(0(), X) -> X , dbl(0()) -> 0() , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [terms](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [0] [recip](x1) = [1] x1 + [0] [sqr](x1) = [1] x1 + [0] [n__terms](x1) = [1] x1 + [4] [n__s](x1) = [1] x1 + [0] [0] = [5] [s](x1) = [1] x1 + [0] [add](x1, x2) = [1] x1 + [1] x2 + [1] [dbl](x1) = [1] x1 + [7] [first](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [5] [n__first](x1, x2) = [1] x1 + [1] x2 + [1] [activate](x1) = [1] x1 + [0] [half](x1) = [1] x1 + [7] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [1] > [1] N + [0] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [1] ? [1] X + [4] = [n__terms(X)] [sqr(0())] = [5] >= [5] = [0()] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [add(0(), X)] = [1] X + [6] > [1] X + [0] = [X] [dbl(0())] = [12] > [5] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [0] ? [1] X1 + [1] X2 + [1] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [5] >= [5] = [nil()] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__terms(X))] = [1] X + [4] > [1] X + [1] = [terms(activate(X))] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(activate(X))] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [1] > [1] X1 + [1] X2 + [0] = [first(activate(X1), activate(X2))] [half(0())] = [12] > [5] = [0()] [half(dbl(X))] = [1] X + [14] > [1] X + [0] = [X] 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 YES(O(1),O(n^1)). Strict Trs: { terms(X) -> n__terms(X) , sqr(0()) -> 0() , s(X) -> n__s(X) , first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) } Weak Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , add(0(), X) -> X , dbl(0()) -> 0() , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [terms](x1) = [1] x1 + [1] [cons](x1, x2) = [1] x1 + [0] [recip](x1) = [1] x1 + [0] [sqr](x1) = [1] x1 + [1] [n__terms](x1) = [1] x1 + [1] [n__s](x1) = [1] x1 + [0] [0] = [7] [s](x1) = [1] x1 + [0] [add](x1, x2) = [1] x1 + [1] x2 + [7] [dbl](x1) = [1] x1 + [7] [first](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [7] [n__first](x1, x2) = [1] x1 + [1] x2 + [0] [activate](x1) = [1] x1 + [0] [half](x1) = [1] x1 + [7] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [1] >= [1] N + [1] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [1] >= [1] X + [1] = [n__terms(X)] [sqr(0())] = [8] > [7] = [0()] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [add(0(), X)] = [1] X + [14] > [1] X + [0] = [X] [dbl(0())] = [14] > [7] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [7] >= [7] = [nil()] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__terms(X))] = [1] X + [1] >= [1] X + [1] = [terms(activate(X))] [activate(n__s(X))] = [1] X + [0] >= [1] X + [0] = [s(activate(X))] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [0] >= [1] X1 + [1] X2 + [0] = [first(activate(X1), activate(X2))] [half(0())] = [14] > [7] = [0()] [half(dbl(X))] = [1] X + [14] > [1] X + [0] = [X] 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 YES(O(1),O(n^1)). Strict Trs: { terms(X) -> n__terms(X) , s(X) -> n__s(X) , first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) } Weak Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , sqr(0()) -> 0() , add(0(), X) -> X , dbl(0()) -> 0() , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following nonconstant growth matrix-interpretation) The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following matrix interpretation satisfying not(EDA) and not(IDA(1)). [terms](x1) = [1] x1 + [0] [cons](x1, x2) = [1] x1 + [0] [recip](x1) = [1] x1 + [0] [sqr](x1) = [1] x1 + [0] [n__terms](x1) = [1] x1 + [0] [n__s](x1) = [1] x1 + [0] [0] = [7] [s](x1) = [1] x1 + [4] [add](x1, x2) = [1] x1 + [1] x2 + [7] [dbl](x1) = [1] x1 + [7] [first](x1, x2) = [1] x1 + [1] x2 + [0] [nil] = [7] [n__first](x1, x2) = [1] x1 + [1] x2 + [4] [activate](x1) = [1] x1 + [0] [half](x1) = [1] x1 + [7] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [0] >= [1] N + [0] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [0] >= [1] X + [0] = [n__terms(X)] [sqr(0())] = [7] >= [7] = [0()] [s(X)] = [1] X + [4] > [1] X + [0] = [n__s(X)] [add(0(), X)] = [1] X + [14] > [1] X + [0] = [X] [dbl(0())] = [14] > [7] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [0] ? [1] X1 + [1] X2 + [4] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [7] >= [7] = [nil()] [activate(X)] = [1] X + [0] >= [1] X + [0] = [X] [activate(n__terms(X))] = [1] X + [0] >= [1] X + [0] = [terms(activate(X))] [activate(n__s(X))] = [1] X + [0] ? [1] X + [4] = [s(activate(X))] [activate(n__first(X1, X2))] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [0] = [first(activate(X1), activate(X2))] [half(0())] = [14] > [7] = [0()] [half(dbl(X))] = [1] X + [14] > [1] X + [0] = [X] 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 YES(O(1),O(n^1)). Strict Trs: { terms(X) -> n__terms(X) , first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) } Weak Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , sqr(0()) -> 0() , s(X) -> n__s(X) , add(0(), X) -> X , dbl(0()) -> 0() , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { terms(X) -> n__terms(X) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [terms](x1) = [1] x1 + [3] [cons](x1, x2) = [1] x1 + [1] [recip](x1) = [1] x1 + [1] [sqr](x1) = [0] [n__terms](x1) = [1] x1 + [2] [n__s](x1) = [1] x1 + [0] [0] = [0] [s](x1) = [1] x1 + [0] [add](x1, x2) = [7] x1 + [7] x2 + [7] [dbl](x1) = [2] x1 + [0] [first](x1, x2) = [1] x1 + [1] x2 + [3] [nil] = [3] [n__first](x1, x2) = [1] x1 + [1] x2 + [3] [activate](x1) = [3] x1 + [0] [half](x1) = [4] x1 + [1] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [3] > [2] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [3] > [1] X + [2] = [n__terms(X)] [sqr(0())] = [0] >= [0] = [0()] [s(X)] = [1] X + [0] >= [1] X + [0] = [n__s(X)] [add(0(), X)] = [7] X + [7] > [1] X + [0] = [X] [dbl(0())] = [0] >= [0] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [3] >= [1] X1 + [1] X2 + [3] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [3] >= [3] = [nil()] [activate(X)] = [3] X + [0] >= [1] X + [0] = [X] [activate(n__terms(X))] = [3] X + [6] > [3] X + [3] = [terms(activate(X))] [activate(n__s(X))] = [3] X + [0] >= [3] X + [0] = [s(activate(X))] [activate(n__first(X1, X2))] = [3] X1 + [3] X2 + [9] > [3] X1 + [3] X2 + [3] = [first(activate(X1), activate(X2))] [half(0())] = [1] > [0] = [0()] [half(dbl(X))] = [8] X + [1] > [1] X + [0] = [X] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) } Weak Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , terms(X) -> n__terms(X) , sqr(0()) -> 0() , s(X) -> n__s(X) , add(0(), X) -> X , dbl(0()) -> 0() , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { first(X1, X2) -> n__first(X1, X2) , activate(n__s(X)) -> s(activate(X)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- The following argument positions are usable: Uargs(terms) = {1}, Uargs(cons) = {1}, Uargs(recip) = {1}, Uargs(s) = {1}, Uargs(first) = {1, 2} TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [terms](x1) = [1] x1 + [4] [cons](x1, x2) = [1] x1 + [0] [recip](x1) = [1] x1 + [0] [sqr](x1) = [4] [n__terms](x1) = [1] x1 + [2] [n__s](x1) = [1] x1 + [2] [0] = [0] [s](x1) = [1] x1 + [2] [add](x1, x2) = [7] x1 + [7] x2 + [7] [dbl](x1) = [1] x1 + [0] [first](x1, x2) = [1] x1 + [1] x2 + [4] [nil] = [3] [n__first](x1, x2) = [1] x1 + [1] x2 + [2] [activate](x1) = [4] x1 + [0] [half](x1) = [2] x1 + [5] The following symbols are considered usable {terms, sqr, s, add, dbl, first, activate, half} The order satisfies the following ordering constraints: [terms(N)] = [1] N + [4] >= [4] = [cons(recip(sqr(N)), n__terms(n__s(N)))] [terms(X)] = [1] X + [4] > [1] X + [2] = [n__terms(X)] [sqr(0())] = [4] > [0] = [0()] [s(X)] = [1] X + [2] >= [1] X + [2] = [n__s(X)] [add(0(), X)] = [7] X + [7] > [1] X + [0] = [X] [dbl(0())] = [0] >= [0] = [0()] [first(X1, X2)] = [1] X1 + [1] X2 + [4] > [1] X1 + [1] X2 + [2] = [n__first(X1, X2)] [first(0(), X)] = [1] X + [4] > [3] = [nil()] [activate(X)] = [4] X + [0] >= [1] X + [0] = [X] [activate(n__terms(X))] = [4] X + [8] > [4] X + [4] = [terms(activate(X))] [activate(n__s(X))] = [4] X + [8] > [4] X + [2] = [s(activate(X))] [activate(n__first(X1, X2))] = [4] X1 + [4] X2 + [8] > [4] X1 + [4] X2 + [4] = [first(activate(X1), activate(X2))] [half(0())] = [5] > [0] = [0()] [half(dbl(X))] = [2] X + [5] > [1] X + [0] = [X] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { terms(N) -> cons(recip(sqr(N)), n__terms(n__s(N))) , terms(X) -> n__terms(X) , sqr(0()) -> 0() , s(X) -> n__s(X) , add(0(), X) -> X , dbl(0()) -> 0() , first(X1, X2) -> n__first(X1, X2) , first(0(), X) -> nil() , activate(X) -> X , activate(n__terms(X)) -> terms(activate(X)) , activate(n__s(X)) -> s(activate(X)) , activate(n__first(X1, X2)) -> first(activate(X1), activate(X2)) , half(0()) -> 0() , half(dbl(X)) -> X } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))