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:
minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)
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(minus1(+2(x, 1)), 1) -> MINUS1(x)
+12(x, minus1(y)) -> MINUS1(+2(minus1(x), y))
+12(x, minus1(y)) -> MINUS1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)
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:
+12(minus1(+2(x, 1)), 1) -> MINUS1(x)
+12(x, minus1(y)) -> MINUS1(+2(minus1(x), y))
+12(x, minus1(y)) -> MINUS1(x)
+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
+12(x, +2(y, z)) -> +12(x, y)
+12(x, minus1(y)) -> +12(minus1(x), y)
+12(x, +2(y, z)) -> +12(+2(x, y), z)
The TRS R consists of the following rules:
minus1(0) -> 0
+2(x, 0) -> x
+2(0, y) -> y
+2(minus1(1), 1) -> 0
minus1(minus1(x)) -> x
+2(x, minus1(y)) -> minus1(+2(minus1(x), y))
+2(x, +2(y, z)) -> +2(+2(x, y), z)
+2(minus1(+2(x, 1)), 1) -> minus1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.