Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup15)

The rewrite relation of the following TRS is considered.

a(a(d(d(x1))))	→	d(d(b(b(x1))))	(1)
a(a(x1))	→	b(b(b(b(b(b(x1))))))	(2)
b(b(d(d(b(b(x1))))))	→	a(a(c(c(x1))))	(3)
c(c(x1))	→	d(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^#(a(d(d(x1))))	→	b^#(b(x1))	(5)
a^#(a(d(d(x1))))	→	b^#(x1)	(6)
a^#(a(x1))	→	b^#(b(b(b(b(b(x1))))))	(7)
a^#(a(x1))	→	b^#(b(b(b(b(x1)))))	(8)
a^#(a(x1))	→	b^#(b(b(b(x1))))	(9)
a^#(a(x1))	→	b^#(b(b(x1)))	(10)
a^#(a(x1))	→	b^#(b(x1))	(11)
a^#(a(x1))	→	b^#(x1)	(12)
b^#(b(d(d(b(b(x1))))))	→	a^#(a(c(c(x1))))	(13)
b^#(b(d(d(b(b(x1))))))	→	a^#(c(c(x1)))	(14)
b^#(b(d(d(b(b(x1))))))	→	c^#(c(x1))	(15)
b^#(b(d(d(b(b(x1))))))	→	c^#(x1)	(16)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

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

1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

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

the pairs

a^#(a(d(d(x1))))	→	b^#(b(x1))	(5)
a^#(a(d(d(x1))))	→	b^#(x1)	(6)

could be deleted.

1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

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

the pair

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

→

a^#(c(c(x1)))

(14)

could be deleted.

1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[a^#(x₁)]	=	-1 + 2 · x₁
[a(x₁)]	=	1 + x₁
[b^#(x₁)]	=	-1 + x₁
[c(x₁)]	=	0
[d(x₁)]	=	-2
[b(x₁)]	=	2

the pair

a^#(a(x1))

→

b^#(x1)

(12)

could be deleted.

1.1.1.1.1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals

[b^#(x₁)]	=	2 · x₁
[b(x₁)]	=	1 + 1 · x₁
[d(x₁)]	=	2 · x₁
[a^#(x₁)]	=	2 · x₁
[a(x₁)]	=	5 + 1 · x₁
[c(x₁)]	=	2 · x₁

the pairs

b^#(b(d(d(b(b(x1))))))	→	a^#(a(c(c(x1))))	(13)
a^#(a(x1))	→	b^#(b(b(b(b(x1)))))	(8)
a^#(a(x1))	→	b^#(b(b(b(x1))))	(9)
a^#(a(x1))	→	b^#(b(b(x1)))	(10)
a^#(a(x1))	→	b^#(b(x1))	(11)

could be deleted.

1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the naturals

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

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

a^#(a(x1))

→

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

(7)

could be deleted.

1.1.1.1.1.1.1.1 P is empty

There are no pairs anymore.