Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup07)

The rewrite relation of the following TRS is considered.

a(a(b(b(x1))))	→	b(b(c(c(a(a(x1))))))	(1)
b(b(c(c(x1))))	→	c(c(b(b(b(b(x1))))))	(2)
a(a(c(c(x1))))	→	c(c(a(a(b(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(b(a(a(x1))))	→	a(a(c(c(b(b(x1))))))	(4)
c(c(b(b(x1))))	→	b(b(b(b(c(c(x1))))))	(5)
c(c(a(a(x1))))	→	b(b(a(a(c(c(x1))))))	(6)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(b(a(a(x1))))	→	b^#(x1)	(7)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(8)
b^#(b(a(a(x1))))	→	c^#(b(b(x1)))	(9)
b^#(b(a(a(x1))))	→	c^#(c(b(b(x1))))	(10)
c^#(c(b(b(x1))))	→	c^#(x1)	(11)
c^#(c(b(b(x1))))	→	c^#(c(x1))	(12)
c^#(c(b(b(x1))))	→	b^#(c(c(x1)))	(13)
c^#(c(b(b(x1))))	→	b^#(b(c(c(x1))))	(14)
c^#(c(b(b(x1))))	→	b^#(b(b(c(c(x1)))))	(15)
c^#(c(b(b(x1))))	→	b^#(b(b(b(c(c(x1))))))	(16)
c^#(c(a(a(x1))))	→	c^#(x1)	(17)
c^#(c(a(a(x1))))	→	c^#(c(x1))	(18)
c^#(c(a(a(x1))))	→	b^#(a(a(c(c(x1)))))	(19)
c^#(c(a(a(x1))))	→	b^#(b(a(a(c(c(x1))))))	(20)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

c^#(c(a(a(x1))))	→	c^#(c(x1))	(18)
c^#(c(a(a(x1))))	→	b^#(b(a(a(c(c(x1))))))	(20)
b^#(b(a(a(x1))))	→	c^#(c(b(b(x1))))	(10)
c^#(c(a(a(x1))))	→	c^#(x1)	(17)
c^#(c(b(b(x1))))	→	b^#(b(b(b(c(c(x1))))))	(16)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(8)
b^#(b(a(a(x1))))	→	b^#(x1)	(7)
c^#(c(b(b(x1))))	→	b^#(b(b(c(c(x1)))))	(15)
c^#(c(b(b(x1))))	→	b^#(b(c(c(x1))))	(14)
c^#(c(b(b(x1))))	→	b^#(c(c(x1)))	(13)
c^#(c(b(b(x1))))	→	c^#(c(x1))	(12)
c^#(c(b(b(x1))))	→	c^#(x1)	(11)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(c^#)	=	1	stat(c^#)	=	lex
prec(b^#)	=	0	stat(b^#)	=	lex
prec(c)	=	0	stat(c)	=	lex
prec(a)	=	6	stat(a)	=	lex
prec(b)	=	0	stat(b)	=	lex

π(c^#)	=	[1]
π(b^#)	=	[1]
π(c)	=	1
π(a)	=	[1]
π(b)	=	1

together with the usable rules

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

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

c^#(c(a(a(x1))))	→	c^#(c(x1))	(18)
c^#(c(a(a(x1))))	→	b^#(b(a(a(c(c(x1))))))	(20)
b^#(b(a(a(x1))))	→	c^#(c(b(b(x1))))	(10)
c^#(c(a(a(x1))))	→	c^#(x1)	(17)
c^#(c(b(b(x1))))	→	b^#(b(b(b(c(c(x1))))))	(16)
b^#(b(a(a(x1))))	→	b^#(b(x1))	(8)
b^#(b(a(a(x1))))	→	b^#(x1)	(7)
c^#(c(b(b(x1))))	→	b^#(b(b(c(c(x1)))))	(15)
c^#(c(b(b(x1))))	→	b^#(b(c(c(x1))))	(14)
c^#(c(b(b(x1))))	→	b^#(c(c(x1)))	(13)

could be deleted.

1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(c^#)	=	0	stat(c^#)	=	lex
prec(c)	=	1	stat(c)	=	lex
prec(a)	=	0	stat(a)	=	lex
prec(b)	=	0	stat(b)	=	lex

π(c^#)	=	1
π(c)	=	[1]
π(a)	=	[]
π(b)	=	1

together with the usable rules

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

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

c^#(c(b(b(x1))))

→

c^#(x1)

(11)

could be deleted.

1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(c^#)	=	0	stat(c^#)	=	lex
prec(c)	=	2	stat(c)	=	lex
prec(a)	=	0	stat(a)	=	lex
prec(b)	=	1	stat(b)	=	lex

π(c^#)	=	[1]
π(c)	=	[1]
π(a)	=	[]
π(b)	=	[1]

together with the usable rules

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

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

c^#(c(b(b(x1))))

→

c^#(c(x1))

(12)

could be deleted.

1.1.1.1.1.1.1 P is empty

There are no pairs anymore.