Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z103)

The rewrite relation of the following TRS is considered.

a(d(x1))	→	d(b(x1))	(1)
a(x1)	→	b(b(b(x1)))	(2)
b(d(b(x1)))	→	a(c(x1))	(3)
c(x1)	→	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))	→	b^#(x1)	(5)
a^#(x1)	→	b^#(b(b(x1)))	(6)
a^#(x1)	→	b^#(b(x1))	(7)
a^#(x1)	→	b^#(x1)	(8)
b^#(d(b(x1)))	→	a^#(c(x1))	(9)
b^#(d(b(x1)))	→	c^#(x1)	(10)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.

The 1^st component contains the pair

b^#(d(b(x1))) → a^#(c(x1)) (9)

a^#(d(x1)) → b^#(x1) (5)

a^#(x1) → b^#(b(b(x1))) (6)

a^#(x1) → b^#(b(x1)) (7)

a^#(x1) → b^#(x1) (8)

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

[a(x₁)] = 1 · x₁

[d(x₁)] = 1 + 1 · x₁

[b(x₁)] = 1 · x₁

[c(x₁)] = 1 + 1 · x₁

[b^#(x₁)] = 1 · x₁

[a^#(x₁)] = 1 · x₁

the pair
a^#(d(x1)) → b^#(x1) (5)
and no rules could be deleted.
1.1.1.1 Monotonic Reduction Pair Processor
Using the linear polynomial interpretation over the naturals

[a(x₁)] = 3 + 1 · x₁

[d(x₁)] = 3 · x₁

[b(x₁)] = 1 + 1 · x₁

[c(x₁)] = 3 · x₁

[b^#(x₁)] = 1 · x₁

[a^#(x₁)] = 3 + 1 · x₁

the pairs

a^#(x1) → b^#(b(b(x1))) (6)

a^#(x1) → b^#(b(x1)) (7)

a^#(x1) → b^#(x1) (8)

and the rule
b(d(b(x1))) → a(c(x1)) (3)
could be deleted.
1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[c(x₁)] = 1 · x₁

[d(x₁)] = 1 · x₁

[b(x₁)] = 1 · x₁

[a^#(x₁)] = 1 · x₁

[b^#(x₁)] = 1 · x₁

together with the usable rule
c(x1) → d(x1) (4)
(w.r.t. the implicit argument filter of the reduction pair), the rule could be deleted.
1.1.1.1.1.1 Switch to Innermost Termination
The TRS does not have overlaps with the pairs and is locally confluent:
20
Hence, it suffices to show innermost termination in the following.
1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals

[b^#(x₁)] = 4 + 5 · x₁

[d(x₁)] = 2 · x₁

[b(x₁)] = 5 · x₁

[a^#(x₁)] = 1

[c(x₁)] = 0

having no usable rules (w.r.t. the implicit argument filter of the reduction pair), the pair
b^#(d(b(x1))) → a^#(c(x1)) (9)
could be deleted.
1.1.1.1.1.1.1.1 P is empty
There are no pairs anymore.

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

[a(x₁)]	=	3 + 1 · x₁
[d(x₁)]	=	3 · x₁
[b(x₁)]	=	1 + 1 · x₁
[c(x₁)]	=	3 · x₁
[b^#(x₁)]	=	1 · x₁
[a^#(x₁)]	=	3 + 1 · x₁

[b^#(x₁)]	=	4 + 5 · x₁
[d(x₁)]	=	2 · x₁
[b(x₁)]	=	5 · x₁
[a^#(x₁)]	=	1
[c(x₁)]	=	0

Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z103)

Property / Task

Answer / Result

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

1.1 Dependency Graph Processor

1.1.1 Monotonic Reduction Pair Processor

1.1.1.1 Monotonic Reduction Pair Processor

1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

1.1.1.1.1.1 Switch to Innermost Termination

1.1.1.1.1.1.1 Reduction Pair Processor with Usable Rules

1.1.1.1.1.1.1.1 P is empty