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:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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:

S(f(x, y)) → F(s(y), s(x))
S(f(x, y)) → S(y)
S(f(x, y)) → S(x)
S(g(x, y)) → G(s(x), s(y))
S(g(x, y)) → S(x)
S(g(x, y)) → S(y)
F(g(x, y), g(u, v)) → G(f(x, u), f(y, v))
F(g(x, y), g(u, v)) → F(x, u)
F(g(x, y), g(u, v)) → F(y, v)

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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:

F(g(x, y), g(u, v)) → F(y, v)
F(g(x, y), g(u, v)) → F(x, u)

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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.


F(g(x, y), g(u, v)) → F(y, v)
F(g(x, y), g(u, v)) → F(x, u)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
F(x0, x1, x2)  =  F(x0)

Tags:
F has argument tags [2,3,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Lexicographic path order with status [LPO].
Quasi-Precedence:
g2 > F2

Status:
F2: [2,1]
g2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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:

S(f(x, y)) → S(x)
S(f(x, y)) → S(y)
S(g(x, y)) → S(x)
S(g(x, y)) → S(y)

The TRS R consists of the following rules:

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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.


S(f(x, y)) → S(x)
S(f(x, y)) → S(y)
S(g(x, y)) → S(x)
S(g(x, y)) → S(y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
S(x0, x1)  =  S(x0)

Tags:
S has argument tags [0,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
S(x1)  =  x1
f(x1, x2)  =  f(x1, x2)
g(x1, x2)  =  g(x1, x2)

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

Status:
f2: [2,1]
g2: [2,1]


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

s(a) → a
s(s(x)) → x
s(f(x, y)) → f(s(y), s(x))
s(g(x, y)) → g(s(x), s(y))
f(x, a) → x
f(a, y) → y
f(g(x, y), g(u, v)) → g(f(x, u), f(y, v))
g(a, a) → a

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