Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-132)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

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

b(a(b(b(x1))))	→	b(b(b(a(x1))))	(4)
b(a(a(a(x1))))	→	a(b(b(a(x1))))	(5)
b(a(a(b(x1))))	→	b(a(a(a(x1))))	(6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(a(b(b(x1))))	→	b^#(a(x1))	(7)
b^#(a(b(b(x1))))	→	b^#(b(a(x1)))	(8)
b^#(a(b(b(x1))))	→	b^#(b(b(a(x1))))	(9)
b^#(a(a(a(x1))))	→	b^#(a(x1))	(10)
b^#(a(a(a(x1))))	→	b^#(b(a(x1)))	(11)
b^#(a(a(b(x1))))	→	b^#(a(a(a(x1))))	(12)

1.1.1 Subterm Criterion Processor

We use the projection to multisets

π(b^#)	=	{ 1, 1 }
π(a)	=	{ 1, 1 }
π(b)	=	{ 1, 1 }

to remove the pairs:

b^#(a(b(b(x1))))	→	b^#(a(x1))	(7)
b^#(a(b(b(x1))))	→	b^#(b(a(x1)))	(8)
b^#(a(a(a(x1))))	→	b^#(a(x1))	(10)
b^#(a(a(a(x1))))	→	b^#(b(a(x1)))	(11)

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair
b^#(a(b(b(x1)))) → b^#(b(b(a(x1)))) (9)

1.1.1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals

[a(x₁)] =

0 1

0 0

· x₁ +

0 0

1 0

[b(x₁)] =

0 0

1 0

· x₁ +

0 0

0 0

[b^#(x₁)] =

0 1

0 0

· x₁ +

2 0

0 0

together with the usable rules

b(a(b(b(x1)))) → b(b(b(a(x1)))) (4)

b(a(a(a(x1)))) → a(b(b(a(x1)))) (5)

b(a(a(b(x1)))) → b(a(a(a(x1)))) (6)

(w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(a(b(b(x1)))) → b^#(b(b(a(x1)))) (9)
could be deleted.
1.1.1.1.1.1 P is empty

There are no pairs anymore.