Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z012)

The rewrite relation of the following TRS is considered.

a(x1)	→	b(x1)	(1)
a(b(x1))	→	b(c(a(x1)))	(2)
b(x1)	→	c(x1)	(3)
c(b(x1))	→	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.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(x1) → c^#(x1) (10)

a^#(b(x1)) → b^#(c(a(x1))) (9)

a^#(b(x1)) → a^#(x1) (8)

c^#(b(x1)) → a^#(x1) (6)

a^#(x1) → b^#(x1) (7)

a^#(b(x1)) → c^#(a(x1)) (5)

1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation

[a(x₁)] = x₁ + 41

[b(x₁)] = x₁ + 41

[c(x₁)] = x₁ + 0

[c^#(x₁)] = x₁ + 0

[a^#(x₁)] = x₁ + 2

[b^#(x₁)] = x₁ + 1

together with the usable rules

c(b(x1)) → a(x1) (4)

a(x1) → b(x1) (1)

b(x1) → c(x1) (3)

a(b(x1)) → b(c(a(x1))) (2)

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

b^#(x1) → c^#(x1) (10)

a^#(b(x1)) → b^#(c(a(x1))) (9)

a^#(b(x1)) → a^#(x1) (8)

c^#(b(x1)) → a^#(x1) (6)

a^#(x1) → b^#(x1) (7)

a^#(b(x1)) → c^#(a(x1)) (5)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.