Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 PisEmptyProof (⇔)
        ↳14 TRUE

(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) 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:

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)))

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(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)))

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)

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(7) 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)))

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:

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)))

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.


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)  =  +
0  =  0

Homeomorphic Embedding Order
The following usable rules [FROCOS05] were oriented: none

(12) 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)))

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