Certification Problem

Input (TPDB SRS_Standard/Waldmann_19/random-317)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

a^#(a(b(a(x1))))	→	b^#(x1)	(5)
a^#(a(b(a(x1))))	→	b^#(b(x1))	(6)
a^#(a(b(a(x1))))	→	b^#(b(b(x1)))	(7)
a^#(a(b(a(x1))))	→	a^#(b(b(b(x1))))	(8)
b^#(b(b(b(x1))))	→	a^#(x1)	(9)
b^#(b(b(b(x1))))	→	a^#(a(x1))	(10)
b^#(b(b(b(x1))))	→	b^#(a(a(x1)))	(11)
b^#(b(b(b(x1))))	→	b^#(b(a(a(x1))))	(12)
a^#(a(b(b(x1))))	→	a^#(x1)	(13)
a^#(a(b(b(x1))))	→	b^#(a(x1))	(14)
a^#(a(b(b(x1))))	→	b^#(b(a(x1)))	(15)
a^#(a(b(b(x1))))	→	a^#(b(b(a(x1))))	(16)
b^#(a(a(a(x1))))	→	b^#(x1)	(17)
b^#(a(a(a(x1))))	→	a^#(b(x1))	(18)
b^#(a(a(a(x1))))	→	b^#(a(b(x1)))	(19)
b^#(a(a(a(x1))))	→	a^#(b(a(b(x1))))	(20)

1.1 Subterm Criterion Processor

We use the projection to multisets

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

to remove the pairs:

a^#(a(b(a(x1))))	→	b^#(x1)	(5)
a^#(a(b(a(x1))))	→	b^#(b(x1))	(6)
a^#(a(b(a(x1))))	→	b^#(b(b(x1)))	(7)
b^#(b(b(b(x1))))	→	a^#(x1)	(9)
b^#(b(b(b(x1))))	→	a^#(a(x1))	(10)
b^#(b(b(b(x1))))	→	b^#(a(a(x1)))	(11)
a^#(a(b(b(x1))))	→	a^#(x1)	(13)
a^#(a(b(b(x1))))	→	b^#(a(x1))	(14)
a^#(a(b(b(x1))))	→	b^#(b(a(x1)))	(15)
b^#(a(a(a(x1))))	→	b^#(x1)	(17)
b^#(a(a(a(x1))))	→	a^#(b(x1))	(18)
b^#(a(a(a(x1))))	→	b^#(a(b(x1)))	(19)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

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

→

b^#(b(a(a(x1))))

(12)

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the

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

π(b^#)	=	[1]
π(b)	=	[1]
π(a)	=	[]

together with the usable rules

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

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

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

→

b^#(b(a(a(x1))))

(12)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pair

a^#(a(b(b(x1))))	→	a^#(b(b(a(x1))))	(16)
a^#(a(b(a(x1))))	→	a^#(b(b(b(x1))))	(8)

1.1.1.2 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

[b(x₁)]

0	0	0
0	0	0
0	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[a(x₁)]

0	0	1
1	0	0
0	0	1

· x₁ +

0	0	0
0	0	0
1	0	0

[a^#(x₁)]

0	0	1
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

together with the usable rules

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

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

a^#(a(b(b(x1))))	→	a^#(b(b(a(x1))))	(16)
a^#(a(b(a(x1))))	→	a^#(b(b(b(x1))))	(8)

could be deleted.

1.1.1.2.1 P is empty

There are no pairs anymore.