Certification Problem

Input (TPDB SRS_Standard/Trafo_06/hom02)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals

[a(x₁)]	=	1 · x₁ + 2
[b(x₁)]	=	1 · x₁ + 1
[c(x₁)]	=	1 · x₁

all of the following rules can be deleted.

b(b(x1))

→

c(c(c(x1)))

(3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

c^#(c(c(b(b(x1))))) → a^#(x1) (9)

a^#(b(b(x1))) → c^#(c(c(a(x1)))) (5)

a^#(b(b(x1))) → a^#(x1) (8)

1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[a(x₁)] = 2 + 1 · x₁

[b(x₁)] = 1 + 1 · x₁

[c(x₁)] = 1 · x₁

[c^#(x₁)] = 1 · x₁

[a^#(x₁)] = 1 + 1 · x₁

the pairs

c^#(c(c(b(b(x1))))) → a^#(x1) (9)

a^#(b(b(x1))) → c^#(c(c(a(x1)))) (5)

a^#(b(b(x1))) → a^#(x1) (8)

and no rules could be deleted.
1.1.1.1.1 P is empty

There are no pairs anymore.