Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-540)

The rewrite relation of the following TRS is considered.

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

[b(x₁)]	=	1 · x₁ + 1
[a(x₁)]	=	1 · x₁

all of the following rules can be deleted.

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

→

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

(1)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
b^#(b(a(a(x1)))) → b^#(x1) (5)

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

[b(x₁)] = 1 · x₁

[a(x₁)] = 1 · x₁

[b^#(x₁)] = 1 · x₁

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
1.1.1.1.1 Size-Change Termination

Using size-change termination in combination with the subterm criterion one obtains the following initial size-change graphs.

b^#(b(a(a(x1)))) → b^#(x1) (5)

1 > 1

As there is no critical graph in the transitive closure, there are no infinite chains.
The 2^nd component contains the pair

b^#(a(a(b(x1)))) → b^#(a(x1)) (7)

b^#(a(a(b(x1)))) → b^#(a(b(a(x1)))) (6)

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

[b(x₁)] = 1 · x₁

[a(x₁)] = 1 · x₁

[b^#(x₁)] = 1 · x₁

together with the usable rule
b(a(a(b(x1)))) → b(a(b(a(x1)))) (3)
(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
  b^#(a(a(b(x1)))) → b^#(a(x1)) (7)
  
  1.1.1.2.1.1 Monotonic Reduction Pair Processor with Usable Rules
  Using the linear polynomial interpretation over the naturals
  
  [a(x₁)] = 1 · x₁
  
  [b(x₁)] = 1 · x₁
  
  [b^#(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
  
  [b^#(x₁)] = -2 + 2 · x₁
  
  [a(x₁)] = x₁
  
  [b(x₁)] = 2 + 2 · x₁
  
  the pair
  b^#(a(a(b(x1)))) → b^#(a(x1)) (7)
  could be deleted.
  1.1.1.2.1.1.1.1 P is empty
  
  There are no pairs anymore.

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

Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-540)

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 with Usable Rules

1.1.1.1.1 Size-Change Termination

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