Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/torpa1)

The rewrite relation of the following TRS is considered.

a(d(x1))	→	d(b(c(b(d(x1)))))	(1)
a(x1)	→	b(b(f(b(b(x1)))))	(2)
b(d(b(x1)))	→	a(d(x1))	(3)
d(f(x1))	→	b(d(x1))	(4)

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.

a^#(d(x1))	→	d^#(b(c(b(d(x1)))))	(5)
a^#(d(x1))	→	b^#(c(b(d(x1))))	(6)
a^#(d(x1))	→	b^#(d(x1))	(7)
a^#(x1)	→	b^#(b(f(b(b(x1)))))	(8)
a^#(x1)	→	b^#(f(b(b(x1))))	(9)
a^#(x1)	→	b^#(b(x1))	(10)
a^#(x1)	→	b^#(x1)	(11)
b^#(d(b(x1)))	→	a^#(d(x1))	(12)
b^#(d(b(x1)))	→	d^#(x1)	(13)
d^#(f(x1))	→	b^#(d(x1))	(14)
d^#(f(x1))	→	d^#(x1)	(15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

a^#(d(x1))	→	b^#(d(x1))	(7)
b^#(d(b(x1)))	→	a^#(d(x1))	(12)
a^#(x1)	→	b^#(b(x1))	(10)
b^#(d(b(x1)))	→	d^#(x1)	(13)
d^#(f(x1))	→	b^#(d(x1))	(14)
d^#(f(x1))	→	d^#(x1)	(15)
a^#(x1)	→	b^#(x1)	(11)

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	1 + 1 · x₁
[d(x₁)]	=	4 · x₁
[b^#(x₁)]	=	1 · x₁
[b(x₁)]	=	1 + 1 · x₁
[d^#(x₁)]	=	4 · x₁
[f(x₁)]	=	1 + 1 · x₁
[a(x₁)]	=	5 + 1 · x₁
[c(x₁)]	=	0

the pairs

a^#(d(x1))	→	b^#(d(x1))	(7)
b^#(d(b(x1)))	→	a^#(d(x1))	(12)
b^#(d(b(x1)))	→	d^#(x1)	(13)
d^#(f(x1))	→	b^#(d(x1))	(14)
d^#(f(x1))	→	d^#(x1)	(15)
a^#(x1)	→	b^#(x1)	(11)

could be deleted.

1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	1
[b^#(x₁)]	=	0
[b(x₁)]	=	0
[d(x₁)]	=	0
[a(x₁)]	=	1
[f(x₁)]	=	1 + 1 · x₁
[c(x₁)]	=	1

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

a^#(x1)

→

b^#(b(x1))

(10)

could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.