Certification Problem

Input (TPDB SRS_Relative/Waldmann_06_relative/r1)

The relative rewrite relation R/S is considered where R is the following TRS

a(b(b(b(a(x1))))) → x1 (1)

and S is the following TRS.

x1 → a(b(a(x1))) (2)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐)}

We obtain the transformed TRS
b(a(b(b(b(a(x1)))))) → b(x1) (3)
a(a(b(b(b(a(x1)))))) → a(x1) (4)
b(x1) → b(a(b(a(x1)))) (5)
a(x1) → a(a(b(a(x1)))) (6)

1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 2):

[b(x1)] = 2x1 + 0
[a(x1)] = 2x1 + 1

We obtain the labeled TRS
a1(a0(b0(b0(b1(a1(x1)))))) → a1(x1) (7)
a1(a0(b0(b0(b1(a0(x1)))))) → a0(x1) (8)
b1(a0(b0(b0(b1(a1(x1)))))) → b1(x1) (9)
b1(a0(b0(b0(b1(a0(x1)))))) → b0(x1) (10)
a1(x1) → a1(a0(b1(a1(x1)))) (11)
a0(x1) → a1(a0(b1(a0(x1)))) (12)
b1(x1) → b1(a0(b1(a1(x1)))) (13)
b0(x1) → b1(a0(b1(a0(x1)))) (14)

1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[b0(x1)] = x1 +
1
[b1(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
0
all of the following rules can be deleted.
a1(a0(b0(b0(b1(a1(x1)))))) → a1(x1) (7)
a1(a0(b0(b0(b1(a0(x1)))))) → a0(x1) (8)
b1(a0(b0(b0(b1(a1(x1)))))) → b1(x1) (9)
b1(a0(b0(b0(b1(a0(x1)))))) → b0(x1) (10)
b0(x1) → b1(a0(b1(a0(x1)))) (14)

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.