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) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(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:

F(s(s(x))) → P(h(g(x)))
F(s(s(x))) → H(g(x))
F(s(s(x))) → G(x)
G(s(x)) → H(g(x))
G(s(x)) → G(x)
H(x) → +1(p(x), q(x))
H(x) → P(x)
H(x) → Q(x)
+1(x, s(y)) → +1(x, y)
F(s(s(x))) → +1(p(g(x)), q(g(x)))
F(s(s(x))) → P(g(x))
F(s(s(x))) → Q(g(x))
G(s(x)) → +1(p(g(x)), q(g(x)))
G(s(x)) → P(g(x))
G(s(x)) → Q(g(x))

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(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 13 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) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(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(x, s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
[f1, g1] > 0 > [s1, p1] > pair2 > +^12
[f1, g1] > 0 > [s1, p1] > q1 > +^12
[f1, g1] > h1 > +2 > [s1, p1] > pair2 > +^12
[f1, g1] > h1 > +2 > [s1, p1] > q1 > +^12


The following usable rules [FROCOS05] were oriented:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

(7) Obligation:

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

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(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:

G(s(x)) → G(x)

The TRS R consists of the following rules:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(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.


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

Recursive Path Order [RPO].
Precedence:
[f1, g1] > 0 > s1 > p1
[f1, g1] > 0 > pair2
[f1, g1] > h1 > pair2
[f1, g1] > h1 > +2 > s1 > p1


The following usable rules [FROCOS05] were oriented:

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(x)))

(12) Obligation:

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

f(0) → 0
f(s(0)) → s(0)
f(s(s(x))) → p(h(g(x)))
g(0) → pair(s(0), s(0))
g(s(x)) → h(g(x))
h(x) → pair(+(p(x), q(x)), p(x))
p(pair(x, y)) → x
q(pair(x, y)) → y
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
f(s(s(x))) → +(p(g(x)), q(g(x)))
g(s(x)) → pair(+(p(g(x)), q(g(x))), p(g(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