(0) Obligation:

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

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

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(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(s(x)) → G(x, s(x))
G(s(x), y) → G(x, +(y, s(x)))
G(s(x), y) → +1(y, s(x))
+1(x, s(y)) → +1(x, y)
G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → +1(y, x)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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:

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(x1))

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

(8) 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)
f(x1)  =  f(x1)
0  =  0
1  =  1
g(x1, x2)  =  g(x1, x2)
+(x1, x2)  =  +(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[f1, 1] > g2 > +2 > s1
0 > s1

Status:
s1: [1]
f1: [1]
0: []
1: []
g2: [1,2]
+2: [2,1]


The following usable rules [FROCOS05] were oriented:

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

(9) Obligation:

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

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) Obligation:

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

G(s(x), y) → G(x, s(+(y, x)))
G(s(x), y) → G(x, +(y, s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(x1))

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


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

Lexicographic path order with status [LPO].
Quasi-Precedence:
[f1, g2] > +2 > s1
[f1, g2] > 1

Status:
s1: [1]
+2: [2,1]
f1: [1]
0: []
1: []
g2: [1,2]


The following usable rules [FROCOS05] were oriented:

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

(14) Obligation:

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

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

The set Q consists of the following terms:

f(0)
f(s(x0))
g(0, x0)
g(s(x0), x1)
+(x0, 0)
+(x0, s(x1))

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE