Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-120)

The rewrite relation of the following TRS is considered.

b(a(b(a(x1))))	→	b(b(a(b(x1))))	(1)
a(a(b(b(x1))))	→	b(b(b(a(x1))))	(2)
b(a(b(b(x1))))	→	b(b(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(b(a(x1))))	→	b^#(b(a(b(x1))))	(4)
b^#(a(b(a(x1))))	→	b^#(a(b(x1)))	(5)
b^#(a(b(a(x1))))	→	a^#(b(x1))	(6)
b^#(a(b(a(x1))))	→	b^#(x1)	(7)
a^#(a(b(b(x1))))	→	b^#(b(b(a(x1))))	(8)
a^#(a(b(b(x1))))	→	b^#(b(a(x1)))	(9)
a^#(a(b(b(x1))))	→	b^#(a(x1))	(10)
a^#(a(b(b(x1))))	→	a^#(x1)	(11)
b^#(a(b(b(x1))))	→	b^#(b(a(a(x1))))	(12)
b^#(a(b(b(x1))))	→	b^#(a(a(x1)))	(13)
b^#(a(b(b(x1))))	→	a^#(a(x1))	(14)
b^#(a(b(b(x1))))	→	a^#(x1)	(15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(a(b(a(x1))))	→	b^#(x1)	(7)
b^#(a(b(a(x1))))	→	b^#(a(b(x1)))	(5)
b^#(a(b(b(x1))))	→	b^#(a(a(x1)))	(13)
b^#(a(b(b(x1))))	→	a^#(a(x1))	(14)
a^#(a(b(b(x1))))	→	b^#(a(x1))	(10)
b^#(a(b(b(x1))))	→	a^#(x1)	(15)
a^#(a(b(b(x1))))	→	a^#(x1)	(11)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

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

the pairs

b^#(a(b(a(x1))))	→	b^#(x1)	(7)
b^#(a(b(a(x1))))	→	b^#(a(b(x1)))	(5)
b^#(a(b(b(x1))))	→	b^#(a(a(x1)))	(13)
b^#(a(b(b(x1))))	→	a^#(a(x1))	(14)
a^#(a(b(b(x1))))	→	b^#(a(x1))	(10)
b^#(a(b(b(x1))))	→	a^#(x1)	(15)
a^#(a(b(b(x1))))	→	a^#(x1)	(11)

could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.