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:

+(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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof

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

F(0, s(0), X) → F(X, +(X, X), X)
+1(X, s(Y)) → +1(X, Y)
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.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
          ↳ QDP

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.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ NonTerminationProof

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.
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:

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


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:



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 [Y' / s(0), X'' / 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 [Y' / s(0), X' / 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.