Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-10)

The rewrite relation of the following TRS is considered.

a(b(a(a(x1))))	→	a(a(a(a(x1))))	(1)
b(a(b(a(x1))))	→	a(a(b(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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(a(b(a(x1))))	→	a^#(b(b(x1)))	(4)
b^#(a(a(b(x1))))	→	a^#(b(a(x1)))	(5)
a^#(b(a(a(x1))))	→	a^#(a(a(a(x1))))	(6)
b^#(a(a(b(x1))))	→	b^#(a(x1))	(7)
a^#(b(a(a(x1))))	→	a^#(a(a(x1)))	(8)
b^#(a(a(b(x1))))	→	b^#(a(b(a(x1))))	(9)
b^#(a(b(a(x1))))	→	b^#(b(x1))	(10)
b^#(a(b(a(x1))))	→	a^#(a(b(b(x1))))	(11)
b^#(a(a(b(x1))))	→	a^#(x1)	(12)
b^#(a(b(a(x1))))	→	b^#(x1)	(13)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

b^#(a(b(a(x1)))) → b^#(x1) (13)

b^#(a(b(a(x1)))) → b^#(b(x1)) (10)

b^#(a(a(b(x1)))) → b^#(a(b(a(x1)))) (9)

b^#(a(a(b(x1)))) → b^#(a(x1)) (7)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[a(x₁)] = x₁ + 1

[b(x₁)] = x₁ + 11293

[a^#(x₁)] = 0

[b^#(x₁)] = x₁ + 0

together with the usable rules

a(b(a(a(x1)))) → a(a(a(a(x1)))) (1)

b(a(a(b(x1)))) → b(a(b(a(x1)))) (3)

b(a(b(a(x1)))) → a(a(b(b(x1)))) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pairs

b^#(a(b(a(x1)))) → b^#(x1) (13)

b^#(a(b(a(x1)))) → b^#(b(x1)) (10)

b^#(a(a(b(x1)))) → b^#(a(x1)) (7)

could be deleted.
1.1.1.1 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(b(a(x1)))) (9)
  
  1.1.1.1.1 Reduction Pair Processor with Usable Rules
  Using the matrix interpretations of dimension 2 with strict dimension 1 over the naturals
  
  [a(x₁)] =
  
  0 1
  
  1 0
  
  · x₁ +
  
  0
  
  0
  
  [b(x₁)] =
  
  1 1
  
  1 1
  
  · x₁ +
  
  2
  
  1
  
  [a^#(x₁)] =
  
  0
  
  0
  
  [b^#(x₁)] =
  
  1 0
  
  1 1
  
  · x₁ +
  
  0
  
  0
  
  together with the usable rules
  
  a(b(a(a(x1)))) → a(a(a(a(x1)))) (1)
  
  b(a(a(b(x1)))) → b(a(b(a(x1)))) (3)
  
  b(a(b(a(x1)))) → a(a(b(b(x1)))) (2)
  
  (w.r.t. the implicit argument filter of the reduction pair), the pair
  b^#(a(a(b(x1)))) → b^#(a(b(a(x1)))) (9)
  could be deleted.
  1.1.1.1.1.1 Dependency Graph Processor
  
  The dependency pairs are split into 0 components.
The 2^nd component contains the pair

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

a^#(b(a(a(x1)))) → a^#(a(a(x1))) (8)

1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[a(x₁)] = 20538

[b(x₁)] = 20539

[a^#(x₁)] = x₁ + 0

[b^#(x₁)] = 0

together with the usable rule
a(b(a(a(x1)))) → a(a(a(a(x1)))) (1)
(w.r.t. the implicit argument filter of the reduction pair), the pairs

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

a^#(b(a(a(x1)))) → a^#(a(a(x1))) (8)

could be deleted.
1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

[a(x₁)]	=	x₁ + 1
[b(x₁)]	=	x₁ + 11293
[a^#(x₁)]	=	0
[b^#(x₁)]	=	x₁ + 0

[a(x₁)]	=	20538
[b(x₁)]	=	20539
[a^#(x₁)]	=	x₁ + 0
[b^#(x₁)]	=	0

Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-10)

Property / Task

Answer / Result

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1 Dependency Graph Processor

1.1.1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1.1.1 Dependency Graph Processor

1.1.2 Reduction Pair Processor with Usable Rules

1.1.2.1 Dependency Graph Processor