Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-95)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

b(c(a(x1)))	→	a(a(c(x1)))	(3)
a(c(c(x1)))	→	c(b(a(x1)))	(4)
b(a(b(x1)))	→	b(c(c(x1)))	(5)
b(b(c(x1)))	→	a(c(b(x1)))	(6)
c(a(a(x1)))	→	c(a(a(x1)))	(7)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(c(a(x1)))	→	a^#(c(x1))	(8)
b^#(c(a(x1)))	→	a^#(a(c(x1)))	(9)
b^#(b(c(x1)))	→	b^#(x1)	(10)
b^#(b(c(x1)))	→	a^#(c(b(x1)))	(11)
b^#(a(b(x1)))	→	b^#(c(c(x1)))	(12)
b^#(a(b(x1)))	→	a^#(x1)	(13)
b^#(a(b(x1)))	→	a^#(a(x1))	(14)
a^#(c(c(x1)))	→	b^#(a(x1))	(15)
a^#(c(c(x1)))	→	a^#(x1)	(16)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1

[c(x₁)]

=

x₁ +

2

[b(x₁)]

=

x₁ +

2

[a(x₁)]

=

x₁ +

2

[b^#(x₁)]

=

x₁ +

2

[a^#(x₁)]

=

x₁ +

1

together with the usable rules

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

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

b^#(c(a(x1)))	→	a^#(c(x1))	(8)
b^#(c(a(x1)))	→	a^#(a(c(x1)))	(9)
b^#(b(c(x1)))	→	b^#(x1)	(10)
b^#(b(c(x1)))	→	a^#(c(b(x1)))	(11)
b^#(a(b(x1)))	→	a^#(x1)	(13)
b^#(a(b(x1)))	→	a^#(a(x1))	(14)
a^#(c(c(x1)))	→	b^#(a(x1))	(15)
a^#(c(c(x1)))	→	a^#(x1)	(16)

and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.