Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-397)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(a(a(b(x1))))	→	b^#(b(b(b(x1))))	(4)
b^#(a(a(b(x1))))	→	b^#(b(b(x1)))	(5)
b^#(a(a(b(x1))))	→	b^#(b(x1))	(6)
b^#(b(b(a(x1))))	→	b^#(b(b(b(x1))))	(7)
b^#(b(b(a(x1))))	→	b^#(b(b(x1)))	(8)
b^#(b(b(a(x1))))	→	b^#(b(x1))	(9)
b^#(b(b(a(x1))))	→	b^#(x1)	(10)
a^#(b(b(b(x1))))	→	b^#(a(a(a(x1))))	(11)
a^#(b(b(b(x1))))	→	a^#(a(a(x1)))	(12)
a^#(b(b(b(x1))))	→	a^#(a(x1))	(13)
a^#(b(b(b(x1))))	→	a^#(x1)	(14)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

a^#(b(b(b(x1)))) → a^#(a(x1)) (13)

a^#(b(b(b(x1)))) → a^#(a(a(x1))) (12)

a^#(b(b(b(x1)))) → a^#(x1) (14)

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 pairs

a^#(b(b(b(x1)))) → a^#(a(x1)) (13)

a^#(b(b(b(x1)))) → a^#(a(a(x1))) (12)

a^#(b(b(b(x1)))) → a^#(x1) (14)

could be deleted.
1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

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

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 rules

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

(w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(b(b(a(x1)))) → b^#(b(x1)) (9)

b^#(b(b(a(x1)))) → b^#(b(b(x1))) (8)

b^#(b(b(a(x1)))) → b^#(x1) (10)

b^#(a(a(b(x1)))) → b^#(b(b(x1))) (5)

b^#(a(a(b(x1)))) → b^#(b(x1)) (6)

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

[b(x₁)] = 1 · x₁

[a(x₁)] = 2 + 1 · x₁

[b^#(x₁)] = 2 · x₁

the pairs

b^#(b(b(a(x1)))) → b^#(b(x1)) (9)

b^#(b(b(a(x1)))) → b^#(b(b(x1))) (8)

b^#(b(b(a(x1)))) → b^#(x1) (10)

b^#(a(a(b(x1)))) → b^#(b(b(x1))) (5)

b^#(a(a(b(x1)))) → b^#(b(x1)) (6)

and the rules

b(b(b(a(x1)))) → b(b(b(b(x1)))) (2)

b(a(a(b(x1)))) → b(b(b(b(x1)))) (1)

could be deleted.
1.1.2.1.1.1 P is empty

There are no pairs anymore.