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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• +¹(s(x), y) → +¹(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 > 1, 2 >= 2

• +¹(x, s(y)) → +¹(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

• +¹(x, +(y, z)) → +¹(+(x, y), z) (allowed arguments on rhs = {2})
  The graph contains the following edges 2 > 2

• +¹(x, +(y, z)) → +¹(x, y) (allowed arguments on rhs = {1, 2})
  The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Combined order from the following AFS and order.
+(x1, x2)  =  +
s(x1)  =  x1
h(x1, x2)  =  h(x2)

Homeomorphic Embedding Order

AFS:
+(x1, x2)  =  +
s(x1)  =  x1
h(x1, x2)  =  h(x2)

From the DPs we obtained the following set of size-change graphs:

• F(h(x, h(y, z))) → F(h(+(x, y), z)) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE