Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-516)

The rewrite relation of the following TRS is considered.

b(b(a(b(x1))))	→	b(b(b(a(x1))))	(1)
b(b(a(a(x1))))	→	a(a(b(b(x1))))	(2)
b(a(a(a(x1))))	→	b(b(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^#(b(a(a(x1))))	→	b^#(x1)	(4)
b^#(b(a(b(x1))))	→	b^#(b(b(a(x1))))	(5)
b^#(b(a(b(x1))))	→	b^#(b(a(x1)))	(6)
b^#(a(a(a(x1))))	→	b^#(a(x1))	(7)
b^#(a(a(a(x1))))	→	b^#(b(b(a(x1))))	(8)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(9)
b^#(b(a(b(x1))))	→	b^#(a(x1))	(10)
b^#(a(a(a(x1))))	→	b^#(b(a(x1)))	(11)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 21240
[b(x₁)]	=	x₁ + 1
[b^#(x₁)]	=	x₁ + 0

together with the usable rules

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

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

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

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^#(b(a(b(x1)))) → b^#(b(b(a(x1)))) (5)

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 0

1 0

· x₁ +

1

1

[b(x₁)] =

0 1

0 0

· x₁ +

2

2

[b^#(x₁)] =

1 1

1 1

· x₁ +

0

0

together with the usable rules

b(b(a(b(x1)))) → b(b(b(a(x1)))) (1)

b(a(a(a(x1)))) → b(b(b(a(x1)))) (3)

b(b(a(a(x1)))) → a(a(b(b(x1)))) (2)

(w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(b(a(b(x1)))) → b^#(b(b(a(x1)))) (5)
could be deleted.
1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.