Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr1)

The rewrite relation of the following TRS is considered.

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

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(x1))	→	b(b(b(b(x1))))	(5)
a(b(x1))	→	a(a(a(a(x1))))	(6)
a(x1)	→	x1	(3)
b(x1)	→	x1	(4)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

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

1.1.1.1 Subterm Criterion Processor

We use the projection to multisets

π(b^#)	=	{ 1 }
π(b)	=	{ 1 }

to remove the pairs:

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

1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

a^#(b(x1))	→	a^#(a(a(a(x1))))	(14)
a^#(b(x1))	→	a^#(x1)	(11)
a^#(b(x1))	→	a^#(a(x1))	(12)
a^#(b(x1))	→	a^#(a(a(x1)))	(13)

1.1.1.2 Subterm Criterion Processor

We use the projection to multisets

π(a^#)	=	{ 1 }
π(a)	=	{ 1 }

to remove the pairs:

a^#(b(x1))	→	a^#(a(a(a(x1))))	(14)
a^#(b(x1))	→	a^#(x1)	(11)
a^#(b(x1))	→	a^#(a(x1))	(12)
a^#(b(x1))	→	a^#(a(a(x1)))	(13)

1.1.1.2.1 P is empty

There are no pairs anymore.