Certification Problem

Input (TPDB SRS_Relative/Waldmann_06_relative/r6)

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

b(b(x1)) → c(b(c(x1))) (1)
c(c(c(x1))) → a(x1) (2)

and S is the following TRS.

a(x1) → a(c(b(x1))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 3 with strict dimension 1 over the naturals
[c(x1)] =
1 0 1
0 0 0
0 2 2
· x1 +
0
0
1
[b(x1)] =
1 2 0
0 4 1
0 4 0
· x1 +
0
2
0
[a(x1)] =
1 6 0
0 0 0
0 0 0
· x1 +
4
0
0
all of the following rules can be deleted.
b(b(x1)) → c(b(c(x1))) (1)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(c(c(c(x1)))) → c(a(x1)) (4)
b(c(c(c(x1)))) → b(a(x1)) (5)
a(c(c(c(x1)))) → a(a(x1)) (6)
c(a(x1)) → c(a(c(b(x1)))) (7)
b(a(x1)) → b(a(c(b(x1)))) (8)
a(a(x1)) → a(a(c(b(x1)))) (9)

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1,2}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 3):

[c(x1)] = 3x1 + 0
[b(x1)] = 3x1 + 1
[a(x1)] = 3x1 + 2

We obtain the labeled TRS
c0(c0(c0(c0(x1)))) → c2(a0(x1)) (10)
c0(c0(c0(c2(x1)))) → c2(a2(x1)) (11)
c0(c0(c0(c1(x1)))) → c2(a1(x1)) (12)
a0(c0(c0(c0(x1)))) → a2(a0(x1)) (13)
a0(c0(c0(c2(x1)))) → a2(a2(x1)) (14)
a0(c0(c0(c1(x1)))) → a2(a1(x1)) (15)
b0(c0(c0(c0(x1)))) → b2(a0(x1)) (16)
b0(c0(c0(c2(x1)))) → b2(a2(x1)) (17)
b0(c0(c0(c1(x1)))) → b2(a1(x1)) (18)
c2(a0(x1)) → c2(a0(c1(b0(x1)))) (19)
c2(a2(x1)) → c2(a0(c1(b2(x1)))) (20)
c2(a1(x1)) → c2(a0(c1(b1(x1)))) (21)
a2(a0(x1)) → a2(a0(c1(b0(x1)))) (22)
a2(a2(x1)) → a2(a0(c1(b2(x1)))) (23)
a2(a1(x1)) → a2(a0(c1(b1(x1)))) (24)
b2(a0(x1)) → b2(a0(c1(b0(x1)))) (25)
b2(a2(x1)) → b2(a0(c1(b2(x1)))) (26)
b2(a1(x1)) → b2(a0(c1(b1(x1)))) (27)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c0(x1)] = x1 +
3/2
[c1(x1)] = x1 +
0
[c2(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b1(x1)] = x1 +
0
[b2(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
1
[a2(x1)] = x1 +
1
all of the following rules can be deleted.
c0(c0(c0(c0(x1)))) → c2(a0(x1)) (10)
c0(c0(c0(c2(x1)))) → c2(a2(x1)) (11)
c0(c0(c0(c1(x1)))) → c2(a1(x1)) (12)
a0(c0(c0(c0(x1)))) → a2(a0(x1)) (13)
a0(c0(c0(c2(x1)))) → a2(a2(x1)) (14)
a0(c0(c0(c1(x1)))) → a2(a1(x1)) (15)
b0(c0(c0(c0(x1)))) → b2(a0(x1)) (16)
b0(c0(c0(c2(x1)))) → b2(a2(x1)) (17)
b0(c0(c0(c1(x1)))) → b2(a1(x1)) (18)
c2(a2(x1)) → c2(a0(c1(b2(x1)))) (20)
c2(a1(x1)) → c2(a0(c1(b1(x1)))) (21)
a2(a2(x1)) → a2(a0(c1(b2(x1)))) (23)
a2(a1(x1)) → a2(a0(c1(b1(x1)))) (24)
b2(a2(x1)) → b2(a0(c1(b2(x1)))) (26)
b2(a1(x1)) → b2(a0(c1(b1(x1)))) (27)

1.1.1.1.1 R is empty

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