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

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

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

(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:

MINUS(s(x), s(y)) → MINUS(x, y)
F(s(x)) → MINUS(s(x), g(f(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)
G(s(x)) → MINUS(s(x), f(g(x)))
G(s(x)) → F(g(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

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 2 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

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.


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

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

Status:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

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:

F(s(x)) → G(f(x))
G(s(x)) → F(g(x))
F(s(x)) → F(x)
G(s(x)) → G(x)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

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

Recursive path order with status [RPO].
Quasi-Precedence:
[s1, f1]

Status:
trivial


The following usable rules [FROCOS05] were oriented:

g(s(x)) → minus(s(x), f(g(x)))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
f(0) → s(0)

(14) Obligation:

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

F(s(x)) → G(f(x))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
f(0) → s(0)
f(s(x)) → minus(s(x), g(f(x)))
g(0) → 0
g(s(x)) → minus(s(x), f(g(x)))

The set Q consists of the following terms:

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

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

(15) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(16) TRUE