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:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), 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:

+¹(x, s(y)) → +¹(x, y)
+¹(s(x), y) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(x, +(y, z)) → +¹(x, y)
F(g(f(x))) → F(h(s(0), x))
F(g(h(x, y))) → F(h(s(x), y))
F(h(x, h(y, z))) → F(h(+(x, y), z))
F(h(x, h(y, z))) → +¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

+¹(s(x), y) → +¹(x, y)
+¹(x, s(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(x, +(y, z)) → +¹(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(s(x), y) → +¹(x, y)
+¹(x, s(y)) → +¹(x, y)
+¹(x, +(y, z)) → +¹(+(x, y), z)
+¹(x, +(y, z)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2)  =  +¹(x1, x2)
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
f(x1)  =  x1
g(x1)  =  g
h(x1, x2)  =  h

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

+^1₂ > +₂ > s₁
g > s₁
g > h

Status:

+^1₂: [2,1]
s₁: [1]
+₂: [2,1]
0: multiset
g: []
h: []

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

(7) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

F(h(x, h(y, z))) → F(h(+(x, y), z))

The TRS R consists of the following rules:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

F(h(x, h(y, z))) → F(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)  =  F(x1)
h(x1, x2)  =  h(x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
f(x1)  =  f
g(x1)  =  g

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

g > [F₁, h₁] > [+₂, s, f]
g > 0 > [+₂, s, f]

Status:

F₁: multiset
h₁: [1]
+₂: [2,1]
0: multiset
s: []
f: []
g: multiset

The following usable rules [FROCOS05] were oriented:

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

(12) Obligation:

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

+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(0, y) → y
+(s(x), y) → s(+(x, y))
+(x, +(y, z)) → +(+(x, y), z)
f(g(f(x))) → f(h(s(0), x))
f(g(h(x, y))) → f(h(s(x), y))
f(h(x, h(y, z))) → f(h(+(x, y), z))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE