Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → 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:
ACTIVATE(n__g(X)) → G(activate(X))
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule ACTIVATE(n__f(X1, X2)) → F(activate(X1), X2) at position [0] we obtained the following new rules:
ACTIVATE(n__f(n__f(x0, x1), y1)) → F(f(activate(x0), x1), y1)
ACTIVATE(n__f(x0, y1)) → F(x0, y1)
ACTIVATE(n__f(n__g(x0), y1)) → F(g(activate(x0)), y1)
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE(n__f(x0, y1)) → F(x0, y1)
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(n__f(x0, x1), y1)) → F(f(activate(x0), x1), y1)
ACTIVATE(n__f(n__g(x0), y1)) → F(g(activate(x0)), y1)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
ACTIVATE(n__f(x0, y1)) → F(x0, y1)
F(g(X), Y) → ACTIVATE(Y)
ACTIVATE(n__f(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(n__f(x0, x1), y1)) → F(f(activate(x0), x1), y1)
ACTIVATE(n__f(n__g(x0), y1)) → F(g(activate(x0)), y1)
F(g(X), Y) → F(X, n__f(n__g(X), activate(Y)))
The TRS R consists of the following rules:
f(g(X), Y) → f(X, n__f(n__g(X), activate(Y)))
f(X1, X2) → n__f(X1, X2)
g(X) → n__g(X)
activate(n__f(X1, X2)) → f(activate(X1), X2)
activate(n__g(X)) → g(activate(X))
activate(X) → X
s = ACTIVATE(n__f(n__g(g(X)), y1)) evaluates to t =ACTIVATE(n__f(n__g(g(X)), activate(y1)))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [y1 / activate(y1)]
Rewriting sequence
ACTIVATE(n__f(n__g(g(X)), y1)) → F(g(activate(g(X))), y1)
with rule ACTIVATE(n__f(n__g(x0), y1')) → F(g(activate(x0)), y1') at position [] and matcher [y1' / y1, x0 / g(X)]
F(g(activate(g(X))), y1) → F(g(g(X)), y1)
with rule activate(X') → X' at position [0,0] and matcher [X' / g(X)]
F(g(g(X)), y1) → F(g(X), n__f(n__g(g(X)), activate(y1)))
with rule F(g(X'), Y') → F(X', n__f(n__g(X'), activate(Y'))) at position [] and matcher [Y' / y1, X' / g(X)]
F(g(X), n__f(n__g(g(X)), activate(y1))) → ACTIVATE(n__f(n__g(g(X)), activate(y1)))
with rule F(g(X), Y) → ACTIVATE(Y)
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.