Termination w.r.t. Q proof of MNZ_10

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

0 QTRS

↳1 DependencyPairsProof (⇔, 9 ms)

↳2 QDP

↳3 DependencyGraphProof (⇔, 0 ms)

↳4 QDP

↳5 TransformationProof (⇔, 0 ms)

↳6 QDP

↳7 DependencyGraphProof (⇔, 0 ms)

↳8 QDP

↳9 QDPOrderProof (⇔, 43 ms)

↳10 QDP

↳11 DependencyGraphProof (⇔, 0 ms)

↳12 AND

    ↳13 QDP

        ↳14 UsableRulesProof (⇔, 0 ms)

        ↳15 QDP

        ↳16 QDPSizeChangeProof (⇔, 0 ms)

        ↳17 YES

    ↳18 QDP

        ↳19 TransformationProof (⇔, 0 ms)

        ↳20 QDP

        ↳21 DependencyGraphProof (⇔, 0 ms)

        ↳22 QDP

        ↳23 TransformationProof (⇔, 0 ms)

        ↳24 QDP

        ↳25 DependencyGraphProof (⇔, 0 ms)

        ↳26 QDP

        ↳27 TransformationProof (⇔, 0 ms)

        ↳28 QDP

        ↳29 DependencyGraphProof (⇔, 0 ms)

        ↳30 QDP

        ↳31 TransformationProof (⇔, 0 ms)

        ↳32 QDP

        ↳33 DependencyGraphProof (⇔, 0 ms)

        ↳34 QDP

        ↳35 TransformationProof (⇔, 0 ms)

        ↳36 QDP

        ↳37 DependencyGraphProof (⇔, 0 ms)

        ↳38 QDP

        ↳39 TransformationProof (⇔, 0 ms)

        ↳40 QDP

        ↳41 DependencyGraphProof (⇔, 0 ms)

        ↳42 QDP

        ↳43 QDPOrderProof (⇔, 26 ms)

        ↳44 QDP

        ↳45 QDPOrderProof (⇔, 503 ms)

        ↳46 QDP

        ↳47 QDPOrderProof (⇔, 687 ms)

        ↳48 QDP

        ↳49 QDPOrderProof (⇔, 411 ms)

        ↳50 QDP

        ↳51 QDPOrderProof (⇔, 398 ms)

        ↳52 QDP

        ↳53 QDPOrderProof (⇔, 400 ms)

        ↳54 QDP

        ↳55 QDPOrderProof (⇔, 439 ms)

        ↳56 QDP

        ↳57 QDPOrderProof (⇔, 723 ms)

        ↳58 QDP

        ↳59 QDPOrderProof (⇔, 1585 ms)

        ↳60 QDP

        ↳61 DependencyGraphProof (⇔, 0 ms)

        ↳62 TRUE

(0) Obligation:

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

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(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:

S(s(0)) → F(0)
S(s(s(0))) → F(s(0))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(s(0))))))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
G(x) → H(x, x)
S(x) → H(x, 0)
S(x) → H(0, x)
F(g(x)) → G(g(f(x)))
F(g(x)) → G(f(x))
F(g(x)) → F(x)
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
H(f(x), g(x)) → F(s(x))
H(f(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(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 1 SCC with 3 less nodes.

(4) Obligation:

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

S(s(s(0))) → F(s(0))
F(s(s(0))) → S(s(s(s(s(s(0))))))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(g(x)) → G(g(f(x)))
G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
F(g(x)) → F(x)
H(f(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(5) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule S(s(s(0))) → F(s(0)) at position [0] we obtained the following new rules [LPAR04]:

S(s(s(0))) → F(h(0, 0)) → S(s(s(0))) → F(h(0, 0))

(6) Obligation:

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

F(s(s(0))) → S(s(s(s(s(s(0))))))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(g(x)) → G(g(f(x)))
G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
F(g(x)) → F(x)
H(f(x), g(x)) → S(x)
S(s(s(0))) → F(h(0, 0))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

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

(8) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(s(0))))))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(g(x)) → G(g(f(x)))
G(x) → H(x, x)
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
G(s(x)) → G(x)
F(g(x)) → F(x)
H(f(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

G(x) → H(x, x)
G(s(x)) → S(s(g(x)))
G(s(x)) → S(g(x))
F(g(x)) → F(x)

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:

POL( F(x₁) ) = x₁

POL( G(x₁) ) = 1

POL( S(x₁) ) = 0

POL( s(x₁) ) = max{0, -1}

POL( h(x₁, x₂) ) = 0

POL( 0 ) = 0

POL( f(x₁) ) = 2x₁

POL( g(x₁) ) = x₁ + 1

POL( H(x₁, x₂) ) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(10) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(s(0))))))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(g(x)) → G(g(f(x)))
H(f(x), g(x)) → F(s(x))
F(g(x)) → G(f(x))
G(s(x)) → G(x)
H(f(x), g(x)) → S(x)

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.

(12) Complex Obligation (AND)

(13) 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:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(14) UsableRulesProof (EQUIVALENT transformation)

We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.

(15) Obligation:

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

G(s(x)) → G(x)

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

(16) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

G(s(x)) → G(x)
The graph contains the following edges 1 > 1

(17) YES

(18) Obligation:

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

F(s(s(0))) → S(s(s(s(s(s(0))))))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(19) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(s(s(s(s(s(0)))))) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(h(s(s(s(s(0)))), 0)) → F(s(s(0))) → S(h(s(s(s(s(0)))), 0))
F(s(s(0))) → S(h(0, s(s(s(s(0)))))) → F(s(s(0))) → S(h(0, s(s(s(s(0))))))
F(s(s(0))) → S(s(h(s(s(s(0))), 0))) → F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0)))))) → F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0))))) → F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0)))) → F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0)))))) → F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0))))) → F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0))))) → F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0)))))) → F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0)))))) → F(s(s(0))) → S(s(s(s(s(h(0, 0))))))

(20) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(s(0)))))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(s(s(0))) → S(h(s(s(s(s(0)))), 0))
F(s(s(0))) → S(h(0, s(s(s(s(0))))))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(21) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(22) Obligation:

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

F(s(s(0))) → S(s(s(s(s(0)))))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(23) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(s(s(s(s(0))))) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(h(s(s(s(0))), 0)) → F(s(s(0))) → S(h(s(s(s(0))), 0))
F(s(s(0))) → S(h(0, s(s(s(0))))) → F(s(s(0))) → S(h(0, s(s(s(0)))))
F(s(s(0))) → S(s(f(s(0)))) → F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0))) → F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0))))) → F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0)))) → F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0)))) → F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0))))) → F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0))))) → F(s(s(0))) → S(s(s(s(h(0, 0)))))

(24) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(s(0))))
F(s(s(0))) → S(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(h(s(s(s(0))), 0))
F(s(s(0))) → S(h(0, s(s(s(0)))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(25) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(26) Obligation:

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

F(s(s(0))) → S(s(s(s(0))))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(27) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(s(s(s(0)))) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(f(s(0))) → F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(h(s(s(0)), 0)) → F(s(s(0))) → S(h(s(s(0)), 0))
F(s(s(0))) → S(h(0, s(s(0)))) → F(s(s(0))) → S(h(0, s(s(0))))
F(s(s(0))) → S(s(f(0))) → F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0))) → F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0)))) → F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0)))) → F(s(s(0))) → S(s(s(h(0, 0))))

(28) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(h(s(s(0)), 0))
F(s(s(0))) → S(h(0, s(s(0))))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(29) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(30) Obligation:

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

F(s(s(0))) → S(s(s(0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(31) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(s(s(0))) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(f(0)) → F(s(s(0))) → S(f(0))
F(s(s(0))) → S(h(s(0), 0)) → F(s(s(0))) → S(h(s(0), 0))
F(s(s(0))) → S(h(0, s(0))) → F(s(s(0))) → S(h(0, s(0)))
F(s(s(0))) → S(s(h(0, 0))) → F(s(s(0))) → S(s(h(0, 0)))

(32) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))
F(s(s(0))) → S(f(0))
F(s(s(0))) → S(h(s(0), 0))
F(s(s(0))) → S(h(0, s(0)))
F(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(33) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(34) Obligation:

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

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))
F(s(s(0))) → S(f(0))
F(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(35) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(f(0)) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(0) → F(s(s(0))) → S(0)

(36) Obligation:

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

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))
F(s(s(0))) → S(s(h(0, 0)))
F(s(s(0))) → S(0)

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(37) DependencyGraphProof (EQUIVALENT transformation)

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

(38) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))
F(s(s(0))) → S(s(h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(39) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule F(s(s(0))) → S(s(h(0, 0))) at position [0] we obtained the following new rules [LPAR04]:

F(s(s(0))) → S(h(h(0, 0), 0)) → F(s(s(0))) → S(h(h(0, 0), 0))
F(s(s(0))) → S(h(0, h(0, 0))) → F(s(s(0))) → S(h(0, h(0, 0)))

(40) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))
F(s(s(0))) → S(h(h(0, 0), 0))
F(s(s(0))) → S(h(0, h(0, 0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(41) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(42) Obligation:

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

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(f(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(43) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(f(0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

POL(F(x₁)) = 0 +
[ 0, 1 ]

· x₁

POL(s(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 1 0 /

· x₁

POL(0) =
/ 0 \

\ 0 /

POL(S(x₁)) = 0 +
[ 0, 1 ]

· x₁

POL(h(x₁, x₂)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 0 0 /

· x₁ +
/ 0 0 \

\ 1 0 /

· x₂

POL(f(x₁)) =
/ 0 \

\ 0 /

+
/ 1 0 \

\ 1 0 /

· x₁

POL(g(x₁)) =
/ 1 \

\ 0 /

+
/ 0 0 \

\ 1 0 /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(44) Obligation:

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

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(h(0, s(0))))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(45) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(h(0, s(0))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(F(x₁)) = 1A +
[ 2A, 3A, 0A ]

· x₁

POL(s(x₁)) =
/ 0A \

| 3A |

\ -I /

+
/ -I 0A 1A \

| 0A 0A 2A |

\ -I -I 1A /

· x₁

POL(0) =
/ 3A \

| -I |

\ -I /

POL(S(x₁)) = 3A +
[ 3A, 0A, 2A ]

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| 0A |

\ -I /

+
/ -I -I -I \

| -I 0A -I |

\ -I -I -I /

· x₁ +
/ -I 0A -I \

| 0A -I 2A |

\ -I -I -I /

· x₂

POL(f(x₁)) =
/ 3A \

| 3A |

\ -I /

+
/ 0A -I 2A \

| 0A -I 2A |

\ -I -I -I /

· x₁

POL(g(x₁)) =
/ 0A \

| 3A |

\ -I /

+
/ 0A 0A 3A \

| 0A 0A 3A |

\ -I -I -I /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(46) Obligation:

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

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(h(s(0), 0)))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(47) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(h(s(s(s(0))), 0)))
F(s(s(0))) → S(s(h(s(s(0)), 0)))
F(s(s(0))) → S(s(h(s(0), 0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(F(x₁)) = 0A +
[ 1A, 0A, -I ]

· x₁

POL(s(x₁)) =
/ 3A \

| 3A |

\ -I /

+
/ 0A -I -I \

| 3A 0A 3A |

\ -I -I 3A /

· x₁

POL(0) =
/ 0A \

| 2A |

\ -I /

POL(S(x₁)) = -I +
[ 0A, 0A, -I ]

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| 3A |

\ -I /

+
/ -I -I -I \

| -I -I -I |

\ -I -I -I /

· x₁ +
/ 0A -I -I \

| 3A -I 0A |

\ -I -I -I /

· x₂

POL(f(x₁)) =
/ -I \

| 3A |

\ -I /

+
/ 0A -I -I \

| 3A -I -I |

\ -I -I -I /

· x₁

POL(g(x₁)) =
/ 3A \

| 3A |

\ -I /

+
/ 0A -I -I \

| 3A -I 0A |

\ -I -I -I /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(48) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(h(0, s(s(0)))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(49) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(h(0, s(s(s(0))))))
F(s(s(0))) → S(s(h(0, s(s(0)))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 3A +
[ -I, -I, 3A ]

· x₁

POL(s(x₁)) =
/ 3A \

| 3A |

\ 3A /

+
/ -I -I -I \

| 0A -I -I |

\ 2A 3A 0A /

· x₁

POL(0) =
/ 1A \

| 0A |

\ 0A /

POL(F(x₁)) = 1A +
[ 0A, -I, 3A ]

· x₁

POL(h(x₁, x₂)) =
/ 3A \

| 1A |

\ -I /

+
/ -I -I -I \

| 0A -I -I |

\ -I 3A -I /

· x₁ +
/ -I -I -I \

| -I -I -I |

\ 2A 0A -I /

· x₂

POL(f(x₁)) =
/ 3A \

| 3A |

\ 0A /

+
/ 0A -I -I \

| 0A 0A -I |

\ 3A 3A -I /

· x₁

POL(g(x₁)) =
/ 3A \

| 3A |

\ 3A /

+
/ 0A 0A -I \

| 0A -I -I |

\ 3A 3A 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(50) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(f(s(0)))
F(s(s(0))) → S(s(s(h(0, 0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(51) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(s(h(s(s(0)), 0))))
F(s(s(0))) → S(s(s(h(0, s(s(0))))))
F(s(s(0))) → S(s(s(s(h(s(0), 0)))))
F(s(s(0))) → S(s(s(s(h(0, s(0))))))
F(s(s(0))) → S(s(s(s(s(h(0, 0))))))
F(s(s(0))) → S(s(s(h(s(0), 0))))
F(s(s(0))) → S(s(s(h(0, s(0)))))
F(s(s(0))) → S(s(s(s(h(0, 0)))))
F(s(s(0))) → S(s(s(h(0, 0))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 0A +
[ 1A, -I, -I ]

· x₁

POL(s(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 0A 0A -I \

| -I 0A -I |

\ -I -I -I /

· x₁

POL(0) =
/ 1A \

| -I |

\ -I /

POL(F(x₁)) = 1A +
[ 1A, -I, -I ]

· x₁

POL(f(x₁)) =
/ 0A \

| 0A |

\ -I /

+
/ 0A -I -I \

| -I 0A -I |

\ -I -I -I /

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| -I |

\ -I /

+
/ -I 0A -I \

| -I 0A -I |

\ -I -I -I /

· x₁ +
/ -I 0A -I \

| -I -I -I |

\ -I -I -I /

· x₂

POL(g(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 0A 0A -I \

| 0A 0A -I |

\ -I -I 2A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(52) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(s(f(0))))
F(s(s(0))) → S(f(s(0)))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(53) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(f(s(0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 0A +
[ -I, 3A, -I ]

· x₁

POL(s(x₁)) =
/ 0A \

| 1A |

\ 0A /

+
/ -I -I -I \

| 0A 0A 3A |

\ 0A -I -I /

· x₁

POL(0) =
/ 0A \

| 0A |

\ -I /

POL(F(x₁)) = 2A +
[ 1A, 3A, -I ]

· x₁

POL(f(x₁)) =
/ 0A \

| 2A |

\ 0A /

+
/ 0A -I -I \

| 1A 0A -I |

\ -I -I -I /

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| -I |

\ 0A /

+
/ -I -I -I \

| -I 0A -I |

\ -I -I -I /

· x₁ +
/ -I -I -I \

| 0A -I 3A |

\ -I -I -I /

· x₂

POL(g(x₁)) =
/ 0A \

| 3A |

\ 0A /

+
/ -I -I 0A \

| 2A 0A 3A |

\ -I -I 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(54) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(f(s(0))))
F(s(s(0))) → S(s(s(f(0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(55) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(f(s(0))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 1A +
[ 1A, -I, 2A ]

· x₁

POL(s(x₁)) =
/ 2A \

| 2A |

\ -I /

+
/ -I -I 2A \

| -I -I 0A |

\ -I 0A 0A /

· x₁

POL(0) =
/ 1A \

| -I |

\ -I /

POL(F(x₁)) = 1A +
[ -I, 3A, -I ]

· x₁

POL(f(x₁)) =
/ 0A \

| 2A |

\ 0A /

+
/ 0A 2A 2A \

| -I 0A 0A |

\ -I -I 0A /

· x₁

POL(h(x₁, x₂)) =
/ 2A \

| 0A |

\ -I /

+
/ -I -I -I \

| -I -I -I |

\ -I -I -I /

· x₁ +
/ -I -I 2A \

| -I -I 0A |

\ -I 0A 0A /

· x₂

POL(g(x₁)) =
/ 3A \

| 2A |

\ 2A /

+
/ 0A 2A 2A \

| -I -I 0A |

\ -I 0A 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(56) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(s(f(0)))))
F(s(s(0))) → S(s(s(f(0))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(57) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

F(s(s(0))) → S(s(s(f(0))))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 2A +
[ 3A, 2A, 0A ]

· x₁

POL(s(x₁)) =
/ 1A \

| -I |

\ -I /

+
/ 0A -I -I \

| -I 0A 2A |

\ 0A -I 0A /

· x₁

POL(0) =
/ 0A \

| -I |

\ -I /

POL(F(x₁)) = 1A +
[ 0A, 3A, -I ]

· x₁

POL(f(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 0A -I 0A \

| -I 1A -I |

\ 0A -I 0A /

· x₁

POL(h(x₁, x₂)) =
/ 1A \

| -I |

\ -I /

+
/ -I -I -I \

| -I 0A 2A |

\ -I -I -I /

· x₁ +
/ 0A -I -I \

| -I 0A 2A |

\ -I -I 0A /

· x₂

POL(g(x₁)) =
/ 1A \

| 3A |

\ 1A /

+
/ 0A -I 0A \

| 3A 0A 3A |

\ 0A -I 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(58) Obligation:

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

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))
F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(s(f(0)))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(59) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].

The following pairs can be oriented strictly and are deleted.

S(s(s(s(s(s(s(s(0)))))))) → F(s(s(0)))

The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:

POL(S(x₁)) = 0A +
[ -I, 3A, 2A ]

· x₁

POL(s(x₁)) =
/ 0A \

| 0A |

\ 1A /

+
/ 0A -I 0A \

| -I -I 0A |

\ 0A 0A -I /

· x₁

POL(0) =
/ 0A \

| 2A |

\ -I /

POL(F(x₁)) = 1A +
[ 1A, 2A, 2A ]

· x₁

POL(f(x₁)) =
/ 0A \

| -I |

\ -I /

+
/ 0A -I -I \

| 0A 0A -I |

\ 0A -I 0A /

· x₁

POL(h(x₁, x₂)) =
/ 0A \

| 0A |

\ 0A /

+
/ -I -I -I \

| -I -I 0A |

\ 0A -I -I /

· x₁ +
/ 0A -I -I \

| -I -I -I |

\ 0A -I -I /

· x₂

POL(g(x₁)) =
/ 1A \

| 0A |

\ 0A /

+
/ 0A 0A 0A \

| 0A -I 0A |

\ 0A 0A 0A /

· x₁

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

s(x) → h(x, 0)
s(x) → h(0, x)
s(s(0)) → f(0)
s(s(s(0))) → f(s(0))
f(s(s(0))) → s(s(s(s(s(s(0))))))
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(g(x)) → g(g(f(x)))
g(x) → h(x, x)
h(f(x), g(x)) → f(s(x))
g(s(x)) → s(s(g(x)))
f(0) → 0
f(s(0)) → s(0)

(60) Obligation:

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

F(s(s(0))) → S(s(s(f(s(0)))))
F(s(s(0))) → S(s(s(s(f(0)))))

The TRS R consists of the following rules:

s(s(0)) → f(0)
f(0) → 0
s(s(s(0))) → f(s(0))
f(s(0)) → s(0)
s(s(s(s(s(s(s(s(0)))))))) → f(s(s(0)))
f(s(s(0))) → s(s(s(s(s(s(0))))))
g(x) → h(x, x)
s(x) → h(x, 0)
s(x) → h(0, x)
f(g(x)) → g(g(f(x)))
g(s(x)) → s(s(g(x)))
h(f(x), g(x)) → f(s(x))

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

(61) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Obligation:

(5) TransformationProof (EQUIVALENT transformation)

(6) Obligation:

(7) DependencyGraphProof (EQUIVALENT transformation)

(8) Obligation:

(9) QDPOrderProof (EQUIVALENT transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Complex Obligation (AND)

(13) Obligation:

(14) UsableRulesProof (EQUIVALENT transformation)

(15) Obligation:

(16) QDPSizeChangeProof (EQUIVALENT transformation)

(17) YES

(18) Obligation:

(19) TransformationProof (EQUIVALENT transformation)

(20) Obligation:

(21) DependencyGraphProof (EQUIVALENT transformation)

(22) Obligation:

(23) TransformationProof (EQUIVALENT transformation)

(24) Obligation:

(25) DependencyGraphProof (EQUIVALENT transformation)

(26) Obligation:

(27) TransformationProof (EQUIVALENT transformation)

(28) Obligation:

(29) DependencyGraphProof (EQUIVALENT transformation)

(30) Obligation:

(31) TransformationProof (EQUIVALENT transformation)

(32) Obligation:

(33) DependencyGraphProof (EQUIVALENT transformation)

(34) Obligation:

(35) TransformationProof (EQUIVALENT transformation)

(36) Obligation:

(37) DependencyGraphProof (EQUIVALENT transformation)

(38) Obligation:

(39) TransformationProof (EQUIVALENT transformation)

(40) Obligation:

(41) DependencyGraphProof (EQUIVALENT transformation)

(42) Obligation:

(43) QDPOrderProof (EQUIVALENT transformation)

(44) Obligation:

(45) QDPOrderProof (EQUIVALENT transformation)

(46) Obligation:

(47) QDPOrderProof (EQUIVALENT transformation)

(48) Obligation:

(49) QDPOrderProof (EQUIVALENT transformation)

(50) Obligation:

(51) QDPOrderProof (EQUIVALENT transformation)

(52) Obligation:

(53) QDPOrderProof (EQUIVALENT transformation)

(54) Obligation:

(55) QDPOrderProof (EQUIVALENT transformation)

(56) Obligation:

(57) QDPOrderProof (EQUIVALENT transformation)

(58) Obligation:

(59) QDPOrderProof (EQUIVALENT transformation)

(60) Obligation:

(61) DependencyGraphProof (EQUIVALENT transformation)

(62) TRUE