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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 UsableRulesProof (⇔)
        ↳7 QDP
        ↳8 QDPSizeChangeProof (⇔)
        ↳9 YES
    ↳10 QDP
        ↳11 NonTerminationProof (⇔)
        ↳12 NO

(0) Obligation:

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(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:

+1(X, s(Y)) → +1(X, Y)
F(0, s(0), X) → F(X, +(X, X), X)
F(0, s(0), X) → +1(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

+1(X, s(Y)) → +1(X, Y)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(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.

(6) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(7) Obligation:

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

+1(X, s(Y)) → +1(X, Y)

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

(8) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • +1(X, s(Y)) → +1(X, Y)
    The graph contains the following edges 1 >= 1, 2 > 2

(9) YES

(10) Obligation:

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

F(0, s(0), X) → F(X, +(X, X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(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.

(11) 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(0, Y), +(g(X', s(0)), g(0, Y')), X) evaluates to t =F(X, +(X, X), X)

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



Rewriting sequence

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

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

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

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

F(0, s(0), g(0, s(0))) → F(g(0, s(0)), +(g(0, s(0)), g(0, s(0))), g(0, s(0)))
with rule F(0, s(0), X) → F(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.



(12) NO