Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/multum3)

The rewrite relation of the following TRS is considered.

a(a(b(x1)))	→	b(b(a(a(x1))))	(1)
b(a(b(x1)))	→	a(a(a(a(x1))))	(2)

Prove or disprove termination.

Yes.

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

b(a(a(x1)))	→	a(a(b(b(x1))))	(3)
b(a(b(x1)))	→	a(a(a(a(x1))))	(2)

The following set of initial dependency pairs has been identified.

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

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

Using the matrix interpretations of dimension 2 with strict dimension 1 over the arctic semiring over the naturals

[b(x₁)]

0	-∞
13	0

· x₁ +

-∞

[a(x₁)]

-∞	2
0	-∞

· x₁ +

[b^#(x₁)]

0	-∞
-∞	-∞

· x₁ +

-∞

together with the usable rules

b(a(a(x1)))	→	a(a(b(b(x1))))	(3)
b(a(b(x1)))	→	a(a(a(a(x1))))	(2)

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

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

could be deleted.

The dependency pairs are split into 0 components.