Termination w.r.t. Q proof of /tmp/tmpejw8OV/21.trs

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 DependencyGraphProof (⇔)

    ↳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:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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:

+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)
+¹(p1, +(p1, x)) → +¹(p2, x)
+¹(p1, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p2, p1) → +¹(p1, p2)
+¹(p2, +(p1, x)) → +¹(p1, +(p2, x))
+¹(p2, +(p1, x)) → +¹(p2, x)
+¹(p2, +(p2, p2)) → +¹(p1, p5)
+¹(p2, +(p2, +(p2, x))) → +¹(p1, +(p5, x))
+¹(p2, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p5, p1) → +¹(p1, p5)
+¹(p5, +(p1, x)) → +¹(p1, +(p5, x))
+¹(p5, +(p1, x)) → +¹(p5, x)
+¹(p5, p2) → +¹(p2, p5)
+¹(p5, +(p2, x)) → +¹(p2, +(p5, x))
+¹(p5, +(p2, x)) → +¹(p5, x)
+¹(p5, +(p5, x)) → +¹(p10, x)
+¹(p10, p1) → +¹(p1, p10)
+¹(p10, +(p1, x)) → +¹(p1, +(p10, x))
+¹(p10, +(p1, x)) → +¹(p10, x)
+¹(p10, p2) → +¹(p2, p10)
+¹(p10, +(p2, x)) → +¹(p2, +(p10, x))
+¹(p10, +(p2, x)) → +¹(p10, x)
+¹(p10, p5) → +¹(p5, p10)
+¹(p10, +(p5, x)) → +¹(p5, +(p10, x))
+¹(p10, +(p5, x)) → +¹(p10, x)

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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 7 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

+¹(p2, +(p1, x)) → +¹(p1, +(p2, x))
+¹(p1, +(p1, x)) → +¹(p2, x)
+¹(p2, +(p1, x)) → +¹(p2, x)
+¹(p2, +(p2, +(p2, x))) → +¹(p1, +(p5, x))
+¹(p1, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p5, +(p1, x)) → +¹(p1, +(p5, x))
+¹(p5, +(p1, x)) → +¹(p5, x)
+¹(p5, +(p2, x)) → +¹(p2, +(p5, x))
+¹(p2, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p5, +(p2, x)) → +¹(p5, x)
+¹(p5, +(p5, x)) → +¹(p10, x)
+¹(p10, +(p1, x)) → +¹(p1, +(p10, x))
+¹(p10, +(p1, x)) → +¹(p10, x)
+¹(p10, +(p2, x)) → +¹(p2, +(p10, x))
+¹(p10, +(p2, x)) → +¹(p10, x)
+¹(p10, +(p5, x)) → +¹(p5, +(p10, x))
+¹(p10, +(p5, x)) → +¹(p10, x)

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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(p1, +(p1, x)) → +¹(p2, x)
+¹(p2, +(p1, x)) → +¹(p2, x)
+¹(p2, +(p2, +(p2, x))) → +¹(p1, +(p5, x))
+¹(p1, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p5, +(p1, x)) → +¹(p5, x)
+¹(p2, +(p2, +(p2, x))) → +¹(p5, x)
+¹(p5, +(p2, x)) → +¹(p5, x)
+¹(p5, +(p5, x)) → +¹(p10, x)
+¹(p10, +(p1, x)) → +¹(p10, x)
+¹(p10, +(p2, x)) → +¹(p10, x)
+¹(p10, +(p5, x)) → +¹(p10, x)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = +¹(x2)
p2 = p2
+(x1, x2) = +(x2)
p1 = p1
p5 = p5
p10 = p10

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

[p2, p1] > p5 > [+^1₁, p10] > +₁

Status:

p1: []
p10: []
p5: []
p2: []
+₁: [1]
+^1₁: [1]

The following usable rules [FROCOS05] were oriented:

+(p10, p2) → +(p2, p10)
+(p10, p5) → +(p5, p10)
+(p10, p1) → +(p1, p10)
+(p5, p2) → +(p2, p5)
+(p5, p1) → +(p1, p5)
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, p1) → +(p1, p2)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p1, +(p1, x)) → +(p2, x)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, +(p5, x)) → +(p5, +(p10, x))
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p5, p5) → p10
+(p1, p1) → p2
+(p1, +(p2, p2)) → p5

(7) Obligation:

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

+¹(p2, +(p1, x)) → +¹(p1, +(p2, x))
+¹(p5, +(p1, x)) → +¹(p1, +(p5, x))
+¹(p5, +(p2, x)) → +¹(p2, +(p5, x))
+¹(p10, +(p1, x)) → +¹(p1, +(p10, x))
+¹(p10, +(p2, x)) → +¹(p2, +(p10, x))
+¹(p10, +(p5, x)) → +¹(p5, +(p10, x))

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, x))

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

(8) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 6 less nodes.

(9) TRUE

(10) Obligation:

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

+¹(+(x, y), z) → +¹(y, z)
+¹(+(x, y), z) → +¹(x, +(y, z))

The TRS R consists of the following rules:

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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.

+¹(+(x, y), z) → +¹(y, z)
+¹(+(x, y), z) → +¹(x, +(y, z))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+¹(x1, x2) = x1
+(x1, x2) = +(x1, x2)
p10 = p10
p1 = p1
p5 = p5
p2 = p2

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

p10 > p5 > [+₂, p1] > p2

Status:

p1: []
p10: []
p5: []
p2: []
+₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

+(p1, p1) → p2
+(p1, +(p2, p2)) → p5
+(p5, p5) → p10
+(+(x, y), z) → +(x, +(y, z))
+(p1, +(p1, x)) → +(p2, x)
+(p1, +(p2, +(p2, x))) → +(p5, x)
+(p2, p1) → +(p1, p2)
+(p2, +(p1, x)) → +(p1, +(p2, x))
+(p2, +(p2, p2)) → +(p1, p5)
+(p2, +(p2, +(p2, x))) → +(p1, +(p5, x))
+(p5, p1) → +(p1, p5)
+(p5, +(p1, x)) → +(p1, +(p5, x))
+(p5, p2) → +(p2, p5)
+(p5, +(p2, x)) → +(p2, +(p5, x))
+(p5, +(p5, x)) → +(p10, x)
+(p10, p1) → +(p1, p10)
+(p10, +(p1, x)) → +(p1, +(p10, x))
+(p10, p2) → +(p2, p10)
+(p10, +(p2, x)) → +(p2, +(p10, x))
+(p10, p5) → +(p5, p10)
+(p10, +(p5, x)) → +(p5, +(p10, 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.

(0) Obligation:

(1) DependencyPairsProof (EQUIVALENT transformation)

(2) Obligation:

(3) DependencyGraphProof (EQUIVALENT transformation)

(4) Complex Obligation (AND)

(5) Obligation:

(6) QDPOrderProof (EQUIVALENT transformation)

(7) Obligation:

(8) DependencyGraphProof (EQUIVALENT transformation)

(9) TRUE

(10) Obligation:

(11) QDPOrderProof (EQUIVALENT transformation)

(12) Obligation:

(13) PisEmptyProof (EQUIVALENT transformation)

(14) TRUE