Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z111)

The rewrite relation of the following TRS is considered.

a(a(x1))	→	b(c(c(c(x1))))	(1)
b(c(x1))	→	d(d(d(d(x1))))	(2)
a(x1)	→	d(c(d(x1)))	(3)
b(b(x1))	→	c(c(c(x1)))	(4)
c(c(x1))	→	d(d(d(x1)))	(5)
c(d(d(x1)))	→	a(x1)	(6)

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^#(x1)	→	c^#(d(x1))	(7)
a^#(a(x1))	→	c^#(x1)	(8)
b^#(b(x1))	→	c^#(x1)	(9)
b^#(b(x1))	→	c^#(c(c(x1)))	(10)
a^#(a(x1))	→	c^#(c(x1))	(11)
a^#(a(x1))	→	b^#(c(c(c(x1))))	(12)
c^#(d(d(x1)))	→	a^#(x1)	(13)
b^#(b(x1))	→	c^#(c(x1))	(14)
a^#(a(x1))	→	c^#(c(c(x1)))	(15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(x1))	→	c^#(c(c(x1)))	(15)
b^#(b(x1))	→	c^#(c(x1))	(14)
c^#(d(d(x1)))	→	a^#(x1)	(13)
a^#(a(x1))	→	b^#(c(c(c(x1))))	(12)
b^#(b(x1))	→	c^#(x1)	(9)
a^#(a(x1))	→	c^#(x1)	(8)
a^#(a(x1))	→	c^#(c(x1))	(11)
b^#(b(x1))	→	c^#(c(c(x1)))	(10)
a^#(x1)	→	c^#(d(x1))	(7)

1.1.1 Reduction Pair Processor with Usable Rules

Using the Max-polynomial interpretation

[a(x₁)]	=	x₁ + 287
[d(x₁)]	=	x₁ + 82
[b(x₁)]	=	x₁ + 205
[c(x₁)]	=	x₁ + 123
[c^#(x₁)]	=	x₁ + 0
[a^#(x₁)]	=	x₁ + 163
[b^#(x₁)]	=	x₁ + 80

together with the usable rules

b(b(x1))	→	c(c(c(x1)))	(4)
a(a(x1))	→	b(c(c(c(x1))))	(1)
a(x1)	→	d(c(d(x1)))	(3)
c(c(x1))	→	d(d(d(x1)))	(5)
c(d(d(x1)))	→	a(x1)	(6)
b(c(x1))	→	d(d(d(d(x1))))	(2)

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

a^#(a(x1))	→	c^#(c(c(x1)))	(15)
b^#(b(x1))	→	c^#(c(x1))	(14)
c^#(d(d(x1)))	→	a^#(x1)	(13)
a^#(a(x1))	→	b^#(c(c(c(x1))))	(12)
b^#(b(x1))	→	c^#(x1)	(9)
a^#(a(x1))	→	c^#(x1)	(8)
a^#(a(x1))	→	c^#(c(x1))	(11)
b^#(b(x1))	→	c^#(c(c(x1)))	(10)
a^#(x1)	→	c^#(d(x1))	(7)

could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.