Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

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)
F(0, s(0), X) → F(X, +(X, X), X)
F(0, s(0), X) → +¹(X, X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

+¹(X, s(Y)) → +¹(X, Y)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

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(X, s(Y)) → +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, x2, x3)  =  f
g(x1, x2)  =  g(x1, x2)

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

[+₂, 0, f] > [+^1₂, s₁]

Status:

+^1₂: [1,2]
s₁: multiset
+₂: [1,2]
0: multiset
f: []
g₂: multiset

The following usable rules [FROCOS05] were oriented:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

(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))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

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(0, s(0), X) → F(X, +(X, X), X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
f(0, s(0), X) → f(X, +(X, X), X)
g(X, Y) → X
g(X, Y) → Y

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