Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z109)

The rewrite relation of the following TRS is considered.

a(a(a(x1)))	→	b(b(x1))	(1)
b(b(b(x1)))	→	c(x1)	(2)
c(x1)	→	d(d(x1))	(3)
d(x1)	→	a(a(x1))	(4)

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^#(x1)	→	d^#(d(x1))	(5)
d^#(x1)	→	a^#(a(x1))	(6)
d^#(x1)	→	a^#(x1)	(7)
a^#(a(a(x1)))	→	b^#(x1)	(8)
c^#(x1)	→	d^#(x1)	(9)
b^#(b(b(x1)))	→	c^#(x1)	(10)
a^#(a(a(x1)))	→	b^#(b(x1))	(11)

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^#(b(x1))	(11)
b^#(b(b(x1)))	→	c^#(x1)	(10)
c^#(x1)	→	d^#(x1)	(9)
d^#(x1)	→	a^#(a(x1))	(6)
a^#(a(a(x1)))	→	b^#(x1)	(8)
d^#(x1)	→	a^#(x1)	(7)
c^#(x1)	→	d^#(d(x1))	(5)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 15446
[d(x₁)]	=	x₁ + 30892
[b(x₁)]	=	x₁ + 23169
[c(x₁)]	=	x₁ + 61784
[d^#(x₁)]	=	x₁ + 15447
[c^#(x₁)]	=	x₁ + 54059
[a^#(x₁)]	=	x₁ + 0
[b^#(x₁)]	=	x₁ + 7722

together with the usable rules

d(x1)	→	a(a(x1))	(4)
a(a(a(x1)))	→	b(b(x1))	(1)
c(x1)	→	d(d(x1))	(3)
b(b(b(x1)))	→	c(x1)	(2)

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

a^#(a(a(x1)))	→	b^#(b(x1))	(11)
b^#(b(b(x1)))	→	c^#(x1)	(10)
c^#(x1)	→	d^#(x1)	(9)
d^#(x1)	→	a^#(a(x1))	(6)
a^#(a(a(x1)))	→	b^#(x1)	(8)
d^#(x1)	→	a^#(x1)	(7)
c^#(x1)	→	d^#(d(x1))	(5)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.