Certification Problem

Input (TPDB TRS_Standard/Secret_06_TRS/5)

The rewrite relation of the following TRS is considered.

a(a(y,0),0)	→	y	(1)
c(c(y))	→	y	(2)
c(a(c(c(y)),x))	→	a(c(c(c(a(x,0)))),y)	(3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals

[a(x₁, x₂)]	=	1 · x₁ + 1 · x₂
[0]	=	0
[c(x₁)]	=	2 + 1 · x₁

all of the following rules can be deleted.

c(c(y))

→

y

(2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

c^#(a(c(c(y)),x))	→	a^#(c(c(c(a(x,0)))),y)	(4)
c^#(a(c(c(y)),x))	→	c^#(c(c(a(x,0))))	(5)
c^#(a(c(c(y)),x))	→	c^#(c(a(x,0)))	(6)
c^#(a(c(c(y)),x))	→	c^#(a(x,0))	(7)
c^#(a(c(c(y)),x))	→	a^#(x,0)	(8)

1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

c^#(a(c(c(y)),x)) → c^#(c(a(x,0))) (6)

c^#(a(c(c(y)),x)) → c^#(c(c(a(x,0)))) (5)

c^#(a(c(c(y)),x)) → c^#(a(x,0)) (7)

1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[a(x₁, x₂)] = 1 · x₁ + 1 · x₂

[0] = 0

[c(x₁)] = 1 + 1 · x₁

[c^#(x₁)] = 1 · x₁

the pairs

c^#(a(c(c(y)),x)) → c^#(c(a(x,0))) (6)

c^#(a(c(c(y)),x)) → c^#(a(x,0)) (7)

and no rules could be deleted.
1.1.1.1.1 Reduction Pair Processor
Using the matrix interpretations of dimension 1 with strict dimension 1 over the arctic semiring over the naturals

[c^#(x₁)] =

-∞

+

1

· x₁

[a(x₁, x₂)] =

-∞

+

1

· x₁ +

4

· x₂

[c(x₁)] =

5

+

1

· x₁

[0] =

0

the pair
c^#(a(c(c(y)),x)) → c^#(c(c(a(x,0)))) (5)
could be deleted.
1.1.1.1.1.1 P is empty

There are no pairs anymore.