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

(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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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

The set Q consists of the following terms:

f(a, g(x0))
f(g(x0), a)
f(g(x0), g(x1))
h(g(x0), x1, x2)
h(a, x0, x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) 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

The set Q consists of the following terms:

f(a, g(x0))
f(g(x0), a)
f(g(x0), g(x1))
h(g(x0), x1, x2)
h(a, x0, x1)

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

(5) 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))
H(g(x), y, z) → F(y, h(x, y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1, x2)  =  x1
g(x1)  =  g(x1)
a  =  a
H(x1, x2, x3)  =  H(x2)
h(x1, x2, x3)  =  h(x1, x2)
f(x1, x2)  =  f

Recursive path order with status [RPO].
Quasi-Precedence:

[g₁, a, H₁, h₂, f]

Status:

a: multiset
f: multiset
g₁: multiset
h₂: multiset
H₁: multiset

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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

F(g(x), g(y)) → H(g(y), x, g(y))
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

The set Q consists of the following terms:

f(a, g(x0))
f(g(x0), a)
f(g(x0), g(x1))
h(g(x0), x1, x2)
h(a, x0, x1)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) 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

The set Q consists of the following terms:

f(a, g(x0))
f(g(x0), a)
f(g(x0), g(x1))
h(g(x0), x1, x2)
h(a, x0, x1)

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

(9) 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: Recursive path order with status [RPO].
Quasi-Precedence:

[H₃, g₁]

Status:

g₁: [1]
H₃: [2,3,1]

The following usable rules [FROCOS05] were oriented: none

(10) 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

The set Q consists of the following terms:

f(a, g(x0))
f(g(x0), a)
f(g(x0), g(x1))
h(g(x0), x1, x2)
h(a, x0, x1)

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

(11) PisEmptyProof (EQUIVALENT transformation)

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

(12) TRUE