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

(0) Obligation:

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

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

-¹(s(x), s(y)) → -¹(x, y)
F(s(x)) → -¹(s(x), g(f(x)))
F(s(x)) → G(f(x))
F(s(x)) → F(x)
G(s(x)) → -¹(s(x), f(g(x)))
G(s(x)) → F(g(x))
G(s(x)) → G(x)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(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 2 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

-¹(s(x), s(y)) → -¹(x, y)

The TRS R consists of the following rules:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(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(s(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(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

-^1₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

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:

-(x, 0) → x
-(0, s(y)) → 0
-(s(x), s(y)) → -(x, y)
f(0) → 0
f(s(x)) → -(s(x), g(f(x)))
g(0) → s(0)
g(s(x)) → -(s(x), f(g(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.

F(s(x)) → G(f(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)  =  g(x1)
0  =  0
-(x1, x2)  =  -(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

[s₁, g₁] > [f₁, -₁] > 0

Status:

s₁: [1]
f₁: [1]
g₁: [1]
0: []
-₁: [1]

The following usable rules [FROCOS05] were oriented:

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

(12) Obligation:

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

G(s(x)) → F(g(x))

The TRS R consists of the following rules:

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

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

(13) DependencyGraphProof (EQUIVALENT transformation)

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

(14) TRUE