Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z025)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 4
[b(x₁)]	=	x₁ + 1
[a^#(x₁)]	=	x₁ + 2
[b^#(x₁)]	=	x₁ + 0

together with the usable rules

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

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

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

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.