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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 QDP
↳7 QDPOrderProof (⇔)
↳8 QDP
↳9 PisEmptyProof (⇔)
↳10 TRUE

(0) Obligation:

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

f(a, g(y)) → g(g(y))
f(g(x), a) → f(x, g(a))
f(g(x), g(y)) → h(g(y), x, g(y))
h(g(x), y, z) → f(y, h(x, y, z))
h(a, y, z) → z

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:

F(g(x), a) → F(x, g(a))
F(g(x), g(y)) → H(g(y), x, g(y))
H(g(x), y, z) → F(y, h(x, y, z))
H(g(x), y, z) → H(x, y, z)

The TRS R consists of the following rules:

f(a, g(y)) → g(g(y))
f(g(x), a) → f(x, g(a))
f(g(x), g(y)) → h(g(y), x, g(y))
h(g(x), y, z) → f(y, h(x, y, z))
h(a, y, z) → z

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(g(x), a) → F(x, g(a))
F(g(x), g(y)) → H(g(y), x, g(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F(x1, x2)) = 1 + x1   
POL(H(x1, x2, x3)) = 1 + x2   
POL(a) = 0   
POL(f(x1, x2)) = 0   
POL(g(x1)) = 1 + x1   
POL(h(x1, x2, x3)) = 0   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

H(g(x), y, z) → F(y, h(x, y, z))
H(g(x), y, z) → H(x, y, z)

The TRS R consists of the following rules:

f(a, g(y)) → g(g(y))
f(g(x), a) → f(x, g(a))
f(g(x), g(y)) → h(g(y), x, g(y))
h(g(x), y, z) → f(y, h(x, y, z))
h(a, y, z) → z

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(6) Obligation:

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

H(g(x), y, z) → H(x, y, z)

The TRS R consists of the following rules:

f(a, g(y)) → g(g(y))
f(g(x), a) → f(x, g(a))
f(g(x), g(y)) → h(g(y), x, g(y))
h(g(x), y, z) → f(y, h(x, y, z))
h(a, y, z) → z

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

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


H(g(x), y, z) → H(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(H(x1, x2, x3)) = x1   
POL(g(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(a, g(y)) → g(g(y))
f(g(x), a) → f(x, g(a))
f(g(x), g(y)) → h(g(y), x, g(y))
h(g(x), y, z) → f(y, h(x, y, z))
h(a, y, z) → z

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

(9) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(10) TRUE