Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/pair3rotate)

The rewrite relation of the following TRS is considered.

p(a(x0),p(b(x1),p(a(x2),x3)))

→

p(x2,p(a(a(x0)),p(b(x1),x3)))

(1)

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.

p^#(a(x0),p(b(x1),p(a(x2),x3)))	→	p^#(b(x1),x3)	(2)
p^#(a(x0),p(b(x1),p(a(x2),x3)))	→	p^#(a(a(x0)),p(b(x1),x3))	(3)
p^#(a(x0),p(b(x1),p(a(x2),x3)))	→	p^#(x2,p(a(a(x0)),p(b(x1),x3)))	(4)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

p^#(a(x0),p(b(x1),p(a(x2),x3)))	→	p^#(a(a(x0)),p(b(x1),x3))	(3)
p^#(a(x0),p(b(x1),p(a(x2),x3)))	→	p^#(x2,p(a(a(x0)),p(b(x1),x3)))	(4)

1.1.1 Reduction Pair Processor with Usable Rules

Using the

prec(p^#)	=	stat(p^#)	=	lex
prec(p)	=	stat(p)	=	lex
prec(b)	=	stat(b)	=	lex
prec(a)	=	stat(a)	=	lex

π(p^#)	=	2
π(p)	=	[2]
π(b)	=	1
π(a)	=	1

together with the usable rule

p(a(x0),p(b(x1),p(a(x2),x3)))

→

p(x2,p(a(a(x0)),p(b(x1),x3)))

(1)

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

p^#(a(x0),p(b(x1),p(a(x2),x3)))

→

p^#(a(a(x0)),p(b(x1),x3))

(3)

could be deleted.

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	0
0	0

· x₁ +

0	0
2	0

[p^#(x₁, x₂)]

2	0
0	0

· x₁ +

0	1
0	0

· x₂ +

2	0
0	0

[a(x₁)]

1	1
0	0

· x₁ +

0	0
0	0

[p(x₁, x₂)]

1	0
2	2

· x₁ +

1	0
2	0

· x₂ +

0	0
0	0

together with the usable rule

p(a(x0),p(b(x1),p(a(x2),x3)))

→

p(x2,p(a(a(x0)),p(b(x1),x3)))

(1)

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

p^#(a(x0),p(b(x1),p(a(x2),x3)))

→

p^#(x2,p(a(a(x0)),p(b(x1),x3)))

(4)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.