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

(0) Obligation:

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

f(0, 1, X) → h(X, X)
h(0, X) → f(0, 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) → H(X, X)
H(0, X) → F(0, X, X)

The TRS R consists of the following rules:

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

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

(3) NonTerminationLoopProof (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 = H(g(X, 0), g(1, Y)) evaluates to t =H(g(1, Y), g(1, Y))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Matcher: [ ]
• Semiunifier: [X / 1, Y / 0]

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Rewriting sequence

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

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

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

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

(4) NO