Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z093)

The rewrite relation of the following TRS is considered.

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

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

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 3
[b(x₁)]	=	x₁ + 1
[c(x₁)]	=	x₁ + 8
[c^#(x₁)]	=	x₁ + 6
[a^#(x₁)]	=	x₁ + 0
[b^#(x₁)]	=	x₁ + 0

together with the usable rules

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

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

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

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.