Termination w.r.t. Q proof of Waldmann_06

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

0 QTRS

↳1 Overlay + Local Confluence (⇔, 0 ms)

↳2 QTRS

↳3 DependencyPairsProof (⇔, 0 ms)

↳4 QDP

↳5 DependencyGraphProof (⇔, 0 ms)

↳6 QDP

↳7 TransformationProof (⇔, 0 ms)

↳8 QDP

↳9 SemLabProof (⇒, 135 ms)

↳10 QDP

↳11 DependencyGraphProof (⇔, 0 ms)

↳12 QDP

↳13 UsableRulesReductionPairsProof (⇔, 0 ms)

↳14 QDP

↳15 DependencyGraphProof (⇔, 0 ms)

↳16 QDP

↳17 QDPOrderProof (⇔, 0 ms)

↳18 QDP

↳19 UsableRulesReductionPairsProof (⇔, 0 ms)

↳20 QDP

↳21 QDPOrderProof (⇔, 4 ms)

↳22 QDP

↳23 PisEmptyProof (⇔, 0 ms)

↳24 YES

(0) Obligation:

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

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, x), y) → F(x, f(a, f(h(a), a)))
F(f(a, x), y) → F(a, f(h(a), a))
F(f(a, x), y) → F(h(a), a)

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Obligation:

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

F(f(a, x), y) → F(x, f(a, f(h(a), a)))
F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(7) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule F(f(a, x), y) → F(x, f(a, f(h(a), a))) we obtained the following new rules [LPAR04]:

F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a))) → F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a)))

(8) Obligation:

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

F(f(a, x), y) → F(y, f(x, f(a, f(h(a), a))))
F(f(a, f(a, y_0)), x1) → F(f(a, y_0), f(a, f(h(a), a)))

The TRS R consists of the following rules:

f(f(a, x), y) → f(y, f(x, f(a, f(h(a), a))))

The set Q consists of the following terms:

f(f(a, x0), x1)

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

(9) SemLabProof (SOUND transformation)

We found the following model for the rules of the TRSs R and P. Interpretation over the domain with elements from 0 to 1.
a: 1
f: 0
h: 0
F: 0
By semantic labelling [SEMLAB] we obtain the following labelled QDP problem.

(10) Obligation:

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

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-0(a., x), y) → F.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-1(f.1-1(a., x), y) → F.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-1(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-1(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(11) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 4 less nodes.

(12) Obligation:

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

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(13) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

The following dependency pairs can be deleted:

F.0-0(f.1-1(a., x), y) → F.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The following rules are removed from R:

f.0-1(f.1-0(a., x), y) → f.1-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-0(f.1-1(a., x), y) → f.0-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
f.0-1(f.1-1(a., x), y) → f.1-0(y, f.1-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 0
POL(f.0-0(x₁, x₂)) = x₁ + x₂
POL(f.0-1(x₁, x₂)) = x₁ + x₂
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(f.1-1(x₁, x₂)) = 1 + x₁ + x₂
POL(h.1(x₁)) = x₁

(14) Obligation:

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

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))
F.0-0(f.1-0(a., f.1-1(a., y_0)), x1) → F.0-0(f.1-1(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(15) DependencyGraphProof (EQUIVALENT transformation)

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

(16) Obligation:

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

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))
F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(17) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

F.0-0(f.1-0(a., x), y) → F.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 1
POL(f.0-0(x₁, x₂)) = 0
POL(f.0-1(x₁, x₂)) = x₁
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(h.1(x₁)) = 0

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

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

(18) Obligation:

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

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The TRS R consists of the following rules:

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(19) UsableRulesReductionPairsProof (EQUIVALENT transformation)

f.0-0(f.1-0(a., x), y) → f.0-0(y, f.0-0(x, f.1-0(a., f.0-1(h.1(a.), a.))))

Used ordering: POLO with Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁ + x₂
POL(a.) = 0
POL(f.0-1(x₁, x₂)) = x₁ + x₂
POL(f.1-0(x₁, x₂)) = x₁ + x₂
POL(h.1(x₁)) = x₁

(20) Obligation:

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

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

R is empty.
The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(21) QDPOrderProof (EQUIVALENT transformation)

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

The following pairs can be oriented strictly and are deleted.

F.0-0(f.1-0(a., f.1-0(a., y_0)), x1) → F.0-0(f.1-0(a., y_0), f.1-0(a., f.0-1(h.1(a.), a.)))

The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(F.0-0(x₁, x₂)) = x₁
POL(a.) = 1
POL(f.0-1(x₁, x₂)) = x₁
POL(f.1-0(x₁, x₂)) = 1 + x₁ + x₂
POL(h.1(x₁)) = 0

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
none

(22) Obligation:

Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:

f.0-0(f.1-0(a., x0), x1)
f.0-1(f.1-0(a., x0), x1)
f.0-0(f.1-1(a., x0), x1)
f.0-1(f.1-1(a., x0), x1)

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

(23) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(0) Obligation:

(1) Overlay + Local Confluence (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyPairsProof (EQUIVALENT transformation)

(4) Obligation:

(5) DependencyGraphProof (EQUIVALENT transformation)

(6) Obligation:

(7) TransformationProof (EQUIVALENT transformation)

(8) Obligation:

(9) SemLabProof (SOUND transformation)

(10) Obligation:

(11) DependencyGraphProof (EQUIVALENT transformation)

(12) Obligation:

(13) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(14) Obligation:

(15) DependencyGraphProof (EQUIVALENT transformation)

(16) Obligation:

(17) QDPOrderProof (EQUIVALENT transformation)

(18) Obligation:

(19) UsableRulesReductionPairsProof (EQUIVALENT transformation)

(20) Obligation:

(21) QDPOrderProof (EQUIVALENT transformation)

(22) Obligation:

(23) PisEmptyProof (EQUIVALENT transformation)

(24) YES