Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z031)

The rewrite relation of the following TRS is considered.

b(c(b(c(a(a(x1))))))

→

a(a(a(b(c(b(c(b(c(x1)))))))))

(1)

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(b(c(b(x1))))))

→

c(b(c(b(c(b(a(a(a(x1)))))))))

(2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(c(b(c(b(x1))))))	→	a^#(x1)	(3)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(x1))	(4)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(a(x1)))	(5)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(c(b(c(b(x1))))))	→	a^#(x1)	(3)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(x1))	(4)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(a(x1)))	(5)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 3 with strict dimension 1 over the naturals

[c(x₁)]

=

0	0	0
0	1	0
1	0	1

· x₁ +

0

0

0

[b(x₁)]

=

0	1	0
1	0	1
2	0	0

· x₁ +

1

0

0

[a(x₁)]

=

0	1	0
1	0	0
2	0	0

· x₁ +

0

0

0

[a^#(x₁)]

=

2	2	0
0	0	0
0	0	0

· x₁ +

0

0

0

together with the usable rule

a(a(c(b(c(b(x1))))))

→

c(b(c(b(c(b(a(a(a(x1)))))))))

(2)

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

a^#(a(c(b(c(b(x1))))))	→	a^#(x1)	(3)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(x1))	(4)
a^#(a(c(b(c(b(x1))))))	→	a^#(a(a(x1)))	(5)

could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.