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:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))



↳ QTRS
  ↳ DependencyPairsProof

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

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))


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(x, h(y)) → F(h(a), y)
F(x, h(y)) → F(f(h(a), y), x)

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation

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

F(x, h(y)) → F(h(a), y)
F(x, h(y)) → F(f(h(a), y), x)

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule F(x, h(y)) → F(h(a), y) we obtained the following new rules:

F(x0, h(h(y_1))) → F(h(a), h(y_1))



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ MNOCProof

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

F(x, h(y)) → F(f(h(a), y), x)
F(x0, h(h(y_1))) → F(h(a), h(y_1))

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

The set Q consists of the following terms:

f(x0, h(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ ForwardInstantiation
        ↳ QDP
          ↳ MNOCProof
            ↳ QDP
              ↳ NonTerminationProof

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

F(x, h(y)) → F(f(h(a), y), x)
F(x0, h(h(y_1))) → F(h(a), h(y_1))

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))

Q is empty.
We have to consider all (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(x, h(y)) → F(f(h(a), y), x)
F(x0, h(h(y_1))) → F(h(a), h(y_1))

The TRS R consists of the following rules:

f(x, h(y)) → h(f(f(h(a), y), x))


s = F(x, f(x', f(x'', h(y)))) evaluates to t =F(f(h(a), f(f(h(a), f(f(h(a), y), x'')), x')), x)

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



Rewriting sequence

F(f(h(y'''), f(x''', h(y'))), f(h(y''), f(x'', h(y)))) → F(f(h(y'''), f(x''', h(y'))), f(h(y''), h(f(f(h(a), y), x''))))
with rule f(x'''', h(y0)) → h(f(f(h(a), y0), x'''')) at position [1,1] and matcher [y0 / y, x'''' / x'']

F(f(h(y'''), f(x''', h(y'))), f(h(y''), h(f(f(h(a), y), x'')))) → F(f(h(y'''), f(x''', h(y'))), h(f(f(h(a), f(f(h(a), y), x'')), h(y''))))
with rule f(x', h(y'0)) → h(f(f(h(a), y'0), x')) at position [1] and matcher [y'0 / f(f(h(a), y), x''), x' / h(y'')]

F(f(h(y'''), f(x''', h(y'))), h(f(f(h(a), f(f(h(a), y), x'')), h(y'')))) → F(f(h(a), f(f(h(a), f(f(h(a), y), x'')), h(y''))), f(h(y'''), f(x''', h(y'))))
with rule F(x, h(y)) → F(f(h(a), y), 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.