Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 NonTerminationProof (⇔)
↳6 FALSE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0, 1, X) → f(g(X, X), X, X)
g(X, Y) → X
g(X, Y) → Y

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(0, 1, X) → F(g(X, X), X, X)
F(0, 1, X) → G(X, X)

The TRS R consists of the following rules:

f(0, 1, X) → f(g(X, X), X, X)
g(X, Y) → X
g(X, Y) → Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(0, 1, X) → F(g(X, X), X, X)

The TRS R consists of the following rules:

f(0, 1, X) → f(g(X, X), X, X)
g(X, Y) → X
g(X, Y) → Y

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(5) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by narrowing to the left:

s = F(g(g(0, Y), Y'), g(X', 1), X) evaluates to t =F(g(X, X), X, X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [X' / 0, Y' / g(0, 1), Y / 1, X / g(0, 1)]



Rewriting sequence

F(g(g(0, 1), g(0, 1)), g(0, 1), g(0, 1)) → F(g(g(0, 1), g(0, 1)), 1, g(0, 1))
with rule g(X', Y'') → Y'' at position [1] and matcher [X' / 0, Y'' / 1]

F(g(g(0, 1), g(0, 1)), 1, g(0, 1)) → F(g(0, 1), 1, g(0, 1))
with rule g(X', Y') → X' at position [0] and matcher [X' / g(0, 1), Y' / g(0, 1)]

F(g(0, 1), 1, g(0, 1)) → F(0, 1, g(0, 1))
with rule g(X', Y) → X' at position [0] and matcher [X' / 0, Y / 1]

F(0, 1, g(0, 1)) → F(g(g(0, 1), g(0, 1)), g(0, 1), g(0, 1))
with rule F(0, 1, X) → F(g(X, X), X, X)

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.



(6) FALSE