(0) Obligation:

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

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(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)
DOUBLE(X) → +1(X, X)
F(0, s(0), X) → F(X, double(X), X)
F(0, s(0), X) → DOUBLE(X)

The TRS R consists of the following rules:

+(X, 0) → X
+(X, s(Y)) → s(+(X, Y))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(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 2 less nodes.

(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))
double(X) → +(X, X)
f(0, s(0), X) → f(X, double(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)  =  x2
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
double(x1)  =  double(x1)
f(x1, x2, x3)  =  f
g(x1, x2)  =  g(x1, x2)

Recursive path order with status [RPO].
Precedence:
double1 > +2 > s1

Status:
trivial

The following usable rules [FROCOS05] were oriented:

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

The TRS R consists of the following rules:

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

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