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: Combined order from the following AFS and order.
F(x1, x2)  =  x1
g(x1, x2)  =  g(x1, x2)
s(x1)  =  s(x1)
a  =  a
f(x1, x2)  =  f(x1, x2)

Lexicographic Path Order [LPO].
Precedence:
s1 > a > g2
s1 > f2 > g2


The following usable rules [FROCOS05] were oriented:

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

(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: Combined order from the following AFS and order.
S(x1)  =  x1
f(x1, x2)  =  f(x1, x2)
g(x1, x2)  =  g(x1, x2)
s(x1)  =  s(x1)
a  =  a

Lexicographic Path Order [LPO].
Precedence:
s1 > f2 > [g2, a]


The following usable rules [FROCOS05] were oriented:

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

(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