Problem Transformed CSR 04 Ex15 Luc98 GM

Tool CaT

Execution Time	Unknown
Answer	MAYBE
Input	Transformed CSR 04 Ex15 Luc98 GM

stdout:

MAYBE

Problem:
a__and(true(),X) -> mark(X)
a__and(false(),Y) -> false()
a__if(true(),X,Y) -> mark(X)
a__if(false(),X,Y) -> mark(Y)
a__add(0(),X) -> mark(X)
a__add(s(X),Y) -> s(add(X,Y))
a__first(0(),X) -> nil()
a__first(s(X),cons(Y,Z)) -> cons(Y,first(X,Z))
a__from(X) -> cons(X,from(s(X)))
mark(and(X1,X2)) -> a__and(mark(X1),X2)
mark(if(X1,X2,X3)) -> a__if(mark(X1),X2,X3)
mark(add(X1,X2)) -> a__add(mark(X1),X2)
mark(first(X1,X2)) -> a__first(mark(X1),mark(X2))
mark(from(X)) -> a__from(X)
mark(true()) -> true()
mark(false()) -> false()
mark(0()) -> 0()
mark(s(X)) -> s(X)
mark(nil()) -> nil()
mark(cons(X1,X2)) -> cons(X1,X2)
a__and(X1,X2) -> and(X1,X2)
a__if(X1,X2,X3) -> if(X1,X2,X3)
a__add(X1,X2) -> add(X1,X2)
a__first(X1,X2) -> first(X1,X2)
a__from(X) -> from(X)

Proof:
Open

Tool IRC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc98 GM

stdout:

YES(?,O(n^1))

Tool IRC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc98 GM

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    innermost runtime-complexity with respect to
  Rules:
    {  a__and(true(), X) -> mark(X)
     , a__and(false(), Y) -> false()
     , a__if(true(), X, Y) -> mark(X)
     , a__if(false(), X, Y) -> mark(Y)
     , a__add(0(), X) -> mark(X)
     , a__add(s(X), Y) -> s(add(X, Y))
     , a__first(0(), X) -> nil()
     , a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
     , a__from(X) -> cons(X, from(s(X)))
     , mark(and(X1, X2)) -> a__and(mark(X1), X2)
     , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
     , mark(add(X1, X2)) -> a__add(mark(X1), X2)
     , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
     , mark(from(X)) -> a__from(X)
     , mark(true()) -> true()
     , mark(false()) -> false()
     , mark(0()) -> 0()
     , mark(s(X)) -> s(X)
     , mark(nil()) -> nil()
     , mark(cons(X1, X2)) -> cons(X1, X2)
     , a__and(X1, X2) -> and(X1, X2)
     , a__if(X1, X2, X3) -> if(X1, X2, X3)
     , a__add(X1, X2) -> add(X1, X2)
     , a__first(X1, X2) -> first(X1, X2)
     , a__from(X) -> from(X)}

Proof Output:
  'matrix-interpretation of dimension 1' proved the best result:

  Details:
  --------
    'matrix-interpretation of dimension 1' succeeded with the following output:
     'matrix-interpretation of dimension 1'
     --------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    innermost runtime-complexity with respect to
       Rules:
         {  a__and(true(), X) -> mark(X)
          , a__and(false(), Y) -> false()
          , a__if(true(), X, Y) -> mark(X)
          , a__if(false(), X, Y) -> mark(Y)
          , a__add(0(), X) -> mark(X)
          , a__add(s(X), Y) -> s(add(X, Y))
          , a__first(0(), X) -> nil()
          , a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
          , a__from(X) -> cons(X, from(s(X)))
          , mark(and(X1, X2)) -> a__and(mark(X1), X2)
          , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
          , mark(add(X1, X2)) -> a__add(mark(X1), X2)
          , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
          , mark(from(X)) -> a__from(X)
          , mark(true()) -> true()
          , mark(false()) -> false()
          , mark(0()) -> 0()
          , mark(s(X)) -> s(X)
          , mark(nil()) -> nil()
          , mark(cons(X1, X2)) -> cons(X1, X2)
          , a__and(X1, X2) -> and(X1, X2)
          , a__if(X1, X2, X3) -> if(X1, X2, X3)
          , a__add(X1, X2) -> add(X1, X2)
          , a__first(X1, X2) -> first(X1, X2)
          , a__from(X) -> from(X)}

     Proof Output:
       The following argument positions are usable:
         Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
         Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {},
         Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {},
         Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {},
         Uargs(if) = {}
       We have the following constructor-restricted matrix interpretation:
       Interpretation Functions:
        a__and(x1, x2) = [1] x1 + [4] x2 + [2]
        true() = [2]
        mark(x1) = [4] x1 + [0]
        false() = [1]
        a__if(x1, x2, x3) = [1] x1 + [4] x2 + [4] x3 + [3]
        a__add(x1, x2) = [1] x1 + [4] x2 + [4]
        0() = [2]
        s(x1) = [1] x1 + [2]
        add(x1, x2) = [1] x1 + [1] x2 + [2]
        a__first(x1, x2) = [1] x1 + [1] x2 + [6]
        nil() = [2]
        cons(x1, x2) = [0] x1 + [0] x2 + [2]
        first(x1, x2) = [1] x1 + [1] x2 + [2]
        a__from(x1) = [0] x1 + [5]
        from(x1) = [0] x1 + [2]
        and(x1, x2) = [1] x1 + [1] x2 + [1]
        if(x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2]

Tool RC1

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc98 GM

stdout:

YES(?,O(n^1))

Tool RC2

Execution Time	Unknown
Answer	YES(?,O(n^1))
Input	Transformed CSR 04 Ex15 Luc98 GM

stdout:

YES(?,O(n^1))

'Fastest (timeout of 60.0 seconds)'
-----------------------------------
Answer:           YES(?,O(n^1))
Input Problem:    runtime-complexity with respect to
  Rules:
    {  a__and(true(), X) -> mark(X)
     , a__and(false(), Y) -> false()
     , a__if(true(), X, Y) -> mark(X)
     , a__if(false(), X, Y) -> mark(Y)
     , a__add(0(), X) -> mark(X)
     , a__add(s(X), Y) -> s(add(X, Y))
     , a__first(0(), X) -> nil()
     , a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
     , a__from(X) -> cons(X, from(s(X)))
     , mark(and(X1, X2)) -> a__and(mark(X1), X2)
     , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
     , mark(add(X1, X2)) -> a__add(mark(X1), X2)
     , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
     , mark(from(X)) -> a__from(X)
     , mark(true()) -> true()
     , mark(false()) -> false()
     , mark(0()) -> 0()
     , mark(s(X)) -> s(X)
     , mark(nil()) -> nil()
     , mark(cons(X1, X2)) -> cons(X1, X2)
     , a__and(X1, X2) -> and(X1, X2)
     , a__if(X1, X2, X3) -> if(X1, X2, X3)
     , a__add(X1, X2) -> add(X1, X2)
     , a__first(X1, X2) -> first(X1, X2)
     , a__from(X) -> from(X)}

Proof Output:
  'matrix-interpretation of dimension 1' proved the best result:

  Details:
  --------
    'matrix-interpretation of dimension 1' succeeded with the following output:
     'matrix-interpretation of dimension 1'
     --------------------------------------
     Answer:           YES(?,O(n^1))
     Input Problem:    runtime-complexity with respect to
       Rules:
         {  a__and(true(), X) -> mark(X)
          , a__and(false(), Y) -> false()
          , a__if(true(), X, Y) -> mark(X)
          , a__if(false(), X, Y) -> mark(Y)
          , a__add(0(), X) -> mark(X)
          , a__add(s(X), Y) -> s(add(X, Y))
          , a__first(0(), X) -> nil()
          , a__first(s(X), cons(Y, Z)) -> cons(Y, first(X, Z))
          , a__from(X) -> cons(X, from(s(X)))
          , mark(and(X1, X2)) -> a__and(mark(X1), X2)
          , mark(if(X1, X2, X3)) -> a__if(mark(X1), X2, X3)
          , mark(add(X1, X2)) -> a__add(mark(X1), X2)
          , mark(first(X1, X2)) -> a__first(mark(X1), mark(X2))
          , mark(from(X)) -> a__from(X)
          , mark(true()) -> true()
          , mark(false()) -> false()
          , mark(0()) -> 0()
          , mark(s(X)) -> s(X)
          , mark(nil()) -> nil()
          , mark(cons(X1, X2)) -> cons(X1, X2)
          , a__and(X1, X2) -> and(X1, X2)
          , a__if(X1, X2, X3) -> if(X1, X2, X3)
          , a__add(X1, X2) -> add(X1, X2)
          , a__first(X1, X2) -> first(X1, X2)
          , a__from(X) -> from(X)}

     Proof Output:
       The following argument positions are usable:
         Uargs(a__and) = {1}, Uargs(mark) = {}, Uargs(a__if) = {1},
         Uargs(a__add) = {1}, Uargs(s) = {}, Uargs(add) = {1},
         Uargs(a__first) = {1, 2}, Uargs(cons) = {}, Uargs(first) = {1, 2},
         Uargs(a__from) = {}, Uargs(from) = {}, Uargs(and) = {1},
         Uargs(if) = {1}
       We have the following constructor-restricted matrix interpretation:
       Interpretation Functions:
        a__and(x1, x2) = [1] x1 + [3] x2 + [2]
        true() = [1]
        mark(x1) = [3] x1 + [0]
        false() = [4]
        a__if(x1, x2, x3) = [1] x1 + [3] x2 + [3] x3 + [6]
        a__add(x1, x2) = [1] x1 + [3] x2 + [5]
        0() = [1]
        s(x1) = [0] x1 + [4]
        add(x1, x2) = [1] x1 + [1] x2 + [4]
        a__first(x1, x2) = [1] x1 + [1] x2 + [7]
        nil() = [3]
        cons(x1, x2) = [0] x1 + [0] x2 + [1]
        first(x1, x2) = [1] x1 + [1] x2 + [4]
        a__from(x1) = [0] x1 + [5]
        from(x1) = [0] x1 + [3]
        and(x1, x2) = [1] x1 + [1] x2 + [1]
        if(x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [4]