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(h(x), y) → h(f(y, f(x, h(a))))

The set Q consists of the following terms:

f(h(x0), x1)



↳ QTRS
  ↳ DependencyPairsProof

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

f(h(x), y) → h(f(y, f(x, h(a))))

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

f(h(x), y) → h(f(y, f(x, h(a))))

The set Q consists of the following terms:

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

The TRS R consists of the following rules:

f(h(x), y) → h(f(y, f(x, h(a))))

The set Q consists of the following terms:

f(h(x0), x1)

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

F(h(h(y_0)), x1) → F(h(y_0), h(a))



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

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

F(h(x), y) → F(y, f(x, h(a)))
F(h(h(y_0)), x1) → F(h(y_0), h(a))

The TRS R consists of the following rules:

f(h(x), y) → h(f(y, f(x, h(a))))

The set Q consists of the following terms:

f(h(x0), 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(h(x), y) → F(y, f(x, h(a)))
F(h(h(y_0)), x1) → F(h(y_0), h(a))

The TRS R consists of the following rules:

f(h(x), y) → h(f(y, f(x, h(a))))

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(h(x), y) → F(y, f(x, h(a)))
F(h(h(y_0)), x1) → F(h(y_0), h(a))

The TRS R consists of the following rules:

f(h(x), y) → h(f(y, f(x, h(a))))


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

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



Rewriting sequence

F(f(f(h(x), y''), h(x'')), f(f(h(x'), y'''), h(x'''))) → F(f(h(f(y'', f(x, h(a)))), h(x'')), f(f(h(x'), y'''), h(x''')))
with rule f(h(x0), y'''') → h(f(y'''', f(x0, h(a)))) at position [0,0] and matcher [y'''' / y'', x0 / x]

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

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

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.