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

0 QTRS
↳1 DependencyPairsProof (⇔, 0 ms)
↳2 QDP
↳3 DependencyGraphProof (⇔, 0 ms)
↳4 QDP
↳5 NonTerminationLoopProof (⇔, 0 ms)
↳6 NO

(0) Obligation:

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

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

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:

G(X) → U(h(X), h(X), X)
G(X) → H(X)
F(k(a), k(b), X) → F(X, X, X)

The TRS R consists of the following rules:

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

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 2 less nodes.

(4) Obligation:

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

F(k(a), k(b), X) → F(X, X, X)

The TRS R consists of the following rules:

g(X) → u(h(X), h(X), X)
u(d, c(Y), X) → k(Y)
h(d) → c(a)
h(d) → c(b)
f(k(a), k(b), X) → f(X, X, X)

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

(5) 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 = F(u(d, h(d), X'), u(d, h(d), X''), X) evaluates to t =F(X, X, X)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [X'' / X', X / u(d, h(d), X')]



Rewriting sequence

F(u(d, h(d), X'), u(d, h(d), X'), u(d, h(d), X')) → F(u(d, h(d), X'), u(d, c(b), X'), u(d, h(d), X'))
with rule h(d) → c(b) at position [1,1] and matcher [ ]

F(u(d, h(d), X'), u(d, c(b), X'), u(d, h(d), X')) → F(u(d, h(d), X'), k(b), u(d, h(d), X'))
with rule u(d, c(Y), X'') → k(Y) at position [1] and matcher [Y / b, X'' / X']

F(u(d, h(d), X'), k(b), u(d, h(d), X')) → F(u(d, c(a), X'), k(b), u(d, h(d), X'))
with rule h(d) → c(a) at position [0,1] and matcher [ ]

F(u(d, c(a), X'), k(b), u(d, h(d), X')) → F(k(a), k(b), u(d, h(d), X'))
with rule u(d, c(Y), X'') → k(Y) at position [0] and matcher [Y / a, X'' / X']

F(k(a), k(b), u(d, h(d), X')) → F(u(d, h(d), X'), u(d, h(d), X'), u(d, h(d), X'))
with rule F(k(a), k(b), X) → F(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) NO