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:
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(x, y)), z)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+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:
+12(f1(x), f1(y)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(f1(x), +2(f1(y), z)) -> +12(f1(+2(x, y)), z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(x, y)), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(f1(x), f1(y)) -> +12(x, y)
+12(f1(x), +2(f1(y), z)) -> +12(x, y)
+12(+2(x, y), z) -> +12(y, z)
+12(f1(x), +2(f1(y), z)) -> +12(f1(+2(x, y)), z)
+12(+2(x, y), z) -> +12(x, +2(y, z))
The TRS R consists of the following rules:
+2(+2(x, y), z) -> +2(x, +2(y, z))
+2(f1(x), f1(y)) -> f1(+2(x, y))
+2(f1(x), +2(f1(y), z)) -> +2(f1(+2(x, y)), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.