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:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
TIMES2(X, s1(Y)) -> TIMES2(Y, X)
TIMES2(X, s1(Y)) -> PLUS2(X, times2(Y, X))
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, 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:
PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
TIMES2(X, s1(Y)) -> TIMES2(Y, X)
TIMES2(X, s1(Y)) -> PLUS2(X, times2(Y, X))
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PLUS2(plus2(X, Y), Z) -> PLUS2(X, plus2(Y, Z))
PLUS2(plus2(X, Y), Z) -> PLUS2(Y, Z)
Used argument filtering: PLUS2(x1, x2) = x1
plus2(x1, x2) = plus2(x1, x2)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
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.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TIMES2(X, s1(Y)) -> TIMES2(Y, X)
The TRS R consists of the following rules:
plus2(plus2(X, Y), Z) -> plus2(X, plus2(Y, Z))
times2(X, s1(Y)) -> plus2(X, times2(Y, X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.