Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-436)

The rewrite relation of the following TRS is considered.

a(b(b(a(x1))))	→	a(b(a(b(x1))))	(1)
a(a(b(a(x1))))	→	a(b(b(a(x1))))	(2)
a(a(b(b(x1))))	→	b(b(a(a(x1))))	(3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals

[a(x₁)]	=	1 · x₁ + 1
[b(x₁)]	=	1 · x₁

all of the following rules can be deleted.

a(a(b(a(x1))))

→

a(b(b(a(x1))))

(2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(b(b(a(x1))))	→	a^#(b(a(b(x1))))	(4)
a^#(b(b(a(x1))))	→	a^#(b(x1))	(5)
a^#(a(b(b(x1))))	→	a^#(a(x1))	(6)
a^#(a(b(b(x1))))	→	a^#(x1)	(7)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

a^#(a(b(b(x1)))) → a^#(x1) (7)

a^#(a(b(b(x1)))) → a^#(a(x1)) (6)

1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[a(x₁)] = 3 + 3 · x₁

[b(x₁)] = 3 + 3 · x₁

[a^#(x₁)] = 2 · x₁

the pairs

a^#(a(b(b(x1)))) → a^#(x1) (7)

a^#(a(b(b(x1)))) → a^#(a(x1)) (6)

and no rules could be deleted.
1.1.1.1.1 P is empty

There are no pairs anymore.
The 2^nd component contains the pair

a^#(b(b(a(x1)))) → a^#(b(x1)) (5)

a^#(b(b(a(x1)))) → a^#(b(a(b(x1)))) (4)

1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[a(x₁)] = 1 · x₁

[b(x₁)] = 1 · x₁

[a^#(x₁)] = 1 · x₁

together with the usable rule
a(b(b(a(x1)))) → a(b(a(b(x1)))) (1)
(w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
1.1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 1 component.
- The 1^st component contains the pair
  a^#(b(b(a(x1)))) → a^#(b(x1)) (5)
  
  1.1.1.2.1.1 Monotonic Reduction Pair Processor with Usable Rules
  Using the linear polynomial interpretation over the naturals
  
  [b(x₁)] = 1 · x₁
  
  [a(x₁)] = 1 · x₁
  
  [a^#(x₁)] = 1 · x₁
  
  having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
  1.1.1.2.1.1.1 Reduction Pair Processor
  Using the linear polynomial interpretation over the naturals
  
  [a^#(x₁)] = 1 · x₁
  
  [b(x₁)] = 1 + 1 · x₁
  
  [a(x₁)] = 1 + 1 · x₁
  
  the pair
  a^#(b(b(a(x1)))) → a^#(b(x1)) (5)
  could be deleted.
  1.1.1.2.1.1.1.1 P is empty
  
  There are no pairs anymore.

[a(x₁)]	=	3 + 3 · x₁
[b(x₁)]	=	3 + 3 · x₁
[a^#(x₁)]	=	2 · x₁

Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-436)

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

1.1 Dependency Pair Transformation

1.1.1 Dependency Graph Processor

1.1.1.1 Monotonic Reduction Pair Processor

1.1.1.1.1 P is empty

1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.2.1 Dependency Graph Processor

1.1.1.2.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.2.1.1.1 Reduction Pair Processor

1.1.1.2.1.1.1.1 P is empty