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 NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

a^#(b(b(b(x1)))) → a^#(a(a(x1))) (5)

a^#(b(b(b(x1)))) → a^#(a(x1)) (8)

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

[a(x₁)] = x₁ + 21239

[b(x₁)] = x₁ + 21239

[a^#(x₁)] = x₁ + 0

[b^#(x₁)] = 0

together with the usable rules

b(a(a(b(x1)))) → b(b(b(b(x1)))) (1)

a(b(b(b(x1)))) → b(a(a(a(x1)))) (3)

b(b(b(a(x1)))) → b(b(b(b(x1)))) (2)

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

a^#(b(b(b(x1)))) → a^#(x1) (13)

a^#(b(b(b(x1)))) → a^#(a(a(x1))) (5)

a^#(b(b(b(x1)))) → a^#(a(x1)) (8)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.

The 2^nd component contains the pair

b^#(b(b(a(x1))))	→	b^#(x1)	(14)
b^#(b(b(a(x1))))	→	b^#(b(b(b(x1))))	(12)
b^#(b(b(a(x1))))	→	b^#(b(x1))	(11)
b^#(a(a(b(x1))))	→	b^#(b(x1))	(7)
b^#(a(a(b(x1))))	→	b^#(b(b(b(x1))))	(6)
b^#(a(a(b(x1))))	→	b^#(b(b(x1)))	(9)
b^#(b(b(a(x1))))	→	b^#(b(b(x1)))	(4)

1.1.2 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 11799
[b(x₁)]	=	x₁ + 11798
[a^#(x₁)]	=	0
[b^#(x₁)]	=	x₁ + 0

together with the usable rules

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

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

b^#(b(b(a(x1))))	→	b^#(x1)	(14)
b^#(b(b(a(x1))))	→	b^#(b(b(b(x1))))	(12)
b^#(b(b(a(x1))))	→	b^#(b(x1))	(11)
b^#(a(a(b(x1))))	→	b^#(b(x1))	(7)
b^#(a(a(b(x1))))	→	b^#(b(b(b(x1))))	(6)
b^#(a(a(b(x1))))	→	b^#(b(b(x1)))	(9)
b^#(b(b(a(x1))))	→	b^#(b(b(x1)))	(4)

could be deleted.

1.1.2.1 Dependency Graph Processor

The dependency pairs are split into 0 components.