Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F1(+2(x, 0)) -> F1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F1(+2(x, 0)) -> F1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(x, +2(y, z)) -> +12(x, y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F1(+2(x, 0)) -> F1(x)
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be strictly oriented and are deleted.
F1(+2(x, 0)) -> F1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1) = F1(x1)
+2(x1, x2) = +1(x1)
0 = 0
Lexicographic Path Order [19].
Precedence:
[F1, +1]
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(+2(x, 0)) -> f1(x)
+2(x, +2(y, z)) -> +2(+2(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.