Certification Problem

Input (TPDB TRS_Standard/Endrullis_06/quadruple1)

The rewrite relation of the following TRS is considered.

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

→

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

(1)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

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

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

→

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

(2)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[p(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[b(x₁)]	=	1 · x₁
[a(x₁)]	=	1 · x₁
[p^#(x₁, x₂)]	=	1 · x₁ + 1 · x₂

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.

1.1.1.1 Reduction Pair Processor

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers

[p^#(x₁, x₂)]

0	1
0	0

· x₁ +

1	0
0	0

· x₂

[p(x₁, x₂)]

0	0
1	1

· x₁ +

1	1
0	0

· x₂

[b(x₁)]

1	0
1	0

· x₁

[a(x₁)]

1	1
0	0

· x₁

the pair

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

→

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

(2)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.