Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/06)

The rewrite relation of the following TRS is considered.

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

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

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(a(x1)))	→	b^#(x1)	(7)
a^#(a(a(x1)))	→	b^#(b(x1))	(8)
a^#(a(a(x1)))	→	a^#(b(b(x1)))	(9)
a^#(b(a(x1)))	→	b^#(x1)	(10)
a^#(b(a(x1)))	→	b^#(b(x1))	(11)
a^#(b(a(x1)))	→	a^#(b(b(x1)))	(12)
b^#(a(b(x1)))	→	a^#(x1)	(13)
b^#(a(b(x1)))	→	a^#(a(x1))	(14)
b^#(a(b(x1)))	→	b^#(a(a(x1)))	(15)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(a(b(x1)))	→	b^#(a(a(x1)))	(15)
b^#(a(b(x1)))	→	a^#(a(x1))	(14)
a^#(a(a(x1)))	→	a^#(b(b(x1)))	(9)
a^#(b(a(x1)))	→	a^#(b(b(x1)))	(12)
a^#(b(a(x1)))	→	b^#(x1)	(10)
b^#(a(b(x1)))	→	a^#(x1)	(13)
a^#(a(a(x1)))	→	b^#(x1)	(7)

1.1.1.1 Subterm Criterion Processor

We use the projection to multisets

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

to remove the pairs:

b^#(a(b(x1)))	→	a^#(a(x1))	(14)
a^#(b(a(x1)))	→	b^#(x1)	(10)
b^#(a(b(x1)))	→	a^#(x1)	(13)
a^#(a(a(x1)))	→	b^#(x1)	(7)

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair
a^#(b(a(x1))) → a^#(b(b(x1))) (12)

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

[b(x₁)] =

0 1

0 0

· x₁ +

0 0

0 0

[a(x₁)] =

0 0

0 0

· x₁ +

0 0

1 0

[a^#(x₁)] =

1 0

0 0

· x₁ +

0 0

0 0

together with the usable rules

a(a(a(x1))) → a(b(b(x1))) (4)

a(b(a(x1))) → a(b(b(x1))) (5)

b(a(b(x1))) → b(a(a(x1))) (6)

(w.r.t. the implicit argument filter of the reduction pair), the pair
a^#(b(a(x1))) → a^#(b(b(x1))) (12)
could be deleted.
1.1.1.1.1.1.1 P is empty

There are no pairs anymore.
The 2^nd component contains the pair
b^#(a(b(x1))) → b^#(a(a(x1))) (15)

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

[b(x₁)] =

0 0

2 0

· x₁ +

0 0

1 0

[b^#(x₁)] =

2 0

0 0

· x₁ +

0 0

0 0

[a(x₁)] =

0 1

2 0

· x₁ +

2 0

0 0

together with the usable rules

a(a(a(x1))) → a(b(b(x1))) (4)

a(b(a(x1))) → a(b(b(x1))) (5)

b(a(b(x1))) → b(a(a(x1))) (6)

(w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(a(b(x1))) → b^#(a(a(x1))) (15)
could be deleted.
1.1.1.1.1.2.1 P is empty

There are no pairs anymore.