Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-76)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

a(a(c(x1)))	→	a(b(c(x1)))	(7)
c(b(c(x1)))	→	b(c(a(x1)))	(8)
c(b(b(x1)))	→	b(b(b(x1)))	(9)
c(b(c(x1)))	→	b(c(b(x1)))	(4)
a(b(b(x1)))	→	c(a(c(x1)))	(10)
a(c(a(x1)))	→	b(b(c(x1)))	(11)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

c^#(b(c(x1)))	→	c^#(b(x1))	(12)
c^#(b(c(x1)))	→	c^#(a(x1))	(13)
c^#(b(c(x1)))	→	a^#(x1)	(14)
a^#(c(a(x1)))	→	c^#(x1)	(15)
a^#(b(b(x1)))	→	c^#(x1)	(16)
a^#(b(b(x1)))	→	c^#(a(c(x1)))	(17)
a^#(b(b(x1)))	→	a^#(c(x1))	(18)
a^#(a(c(x1)))	→	a^#(b(c(x1)))	(19)

1.1.1 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 +

1

[b(x₁)]

=

x₁ +

1

[a(x₁)]

=

x₁ +

1

[c^#(x₁)]

=

x₁ +

0

[a^#(x₁)]

=

x₁ +

1

together with the usable rules

a(a(c(x1)))	→	a(b(c(x1)))	(7)
c(b(c(x1)))	→	b(c(a(x1)))	(8)
c(b(b(x1)))	→	b(b(b(x1)))	(9)
c(b(c(x1)))	→	b(c(b(x1)))	(4)
a(b(b(x1)))	→	c(a(c(x1)))	(10)
a(c(a(x1)))	→	b(b(c(x1)))	(11)

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

c^#(b(c(x1)))	→	c^#(b(x1))	(12)
c^#(b(c(x1)))	→	c^#(a(x1))	(13)
c^#(b(c(x1)))	→	a^#(x1)	(14)
a^#(c(a(x1)))	→	c^#(x1)	(15)
a^#(b(b(x1)))	→	c^#(x1)	(16)
a^#(b(b(x1)))	→	c^#(a(c(x1)))	(17)
a^#(b(b(x1)))	→	a^#(c(x1))	(18)

and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.