Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z106)

The rewrite relation of the following TRS is considered.

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

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.

b^#(x1)	→	c^#(d(x1))	(6)
a^#(a(x1))	→	b^#(x1)	(7)
b^#(c(x1))	→	c^#(b(x1))	(8)
a^#(a(x1))	→	b^#(b(x1))	(9)
b^#(c(d(x1)))	→	a^#(x1)	(10)
b^#(x1)	→	c^#(c(d(x1)))	(11)
b^#(c(x1))	→	b^#(x1)	(12)
a^#(a(x1))	→	b^#(b(b(x1)))	(13)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(a(x1)) → b^#(b(b(x1))) (13)

b^#(c(x1)) → b^#(x1) (12)

b^#(c(d(x1))) → a^#(x1) (10)

a^#(a(x1)) → b^#(x1) (7)

a^#(a(x1)) → b^#(b(x1)) (9)

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

[a(x₁)] = x₁ + 22

[d(x₁)] = x₁ + 2

[b(x₁)] = x₁ + 14

[c(x₁)] = x₁ + 6

[c^#(x₁)] = 0

[a^#(x₁)] = x₁ + 7

[b^#(x₁)] = x₁ + 0

together with the usable rules

b(c(x1)) → c(b(x1)) (4)

a(a(x1)) → b(b(b(x1))) (1)

c(x1) → d(d(d(x1))) (3)

b(c(d(x1))) → a(x1) (5)

b(x1) → c(c(d(x1))) (2)

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

a^#(a(x1)) → b^#(b(b(x1))) (13)

b^#(c(x1)) → b^#(x1) (12)

b^#(c(d(x1))) → a^#(x1) (10)

a^#(a(x1)) → b^#(x1) (7)

a^#(a(x1)) → b^#(b(x1)) (9)

could be deleted.
1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.