Certification Problem

Input (TPDB SRS_Relative/Waldmann_06_relative/r7)

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

b(b(b(x1))) → x1 (1)
c(c(c(x1))) → a(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 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(b(b(b(x1)))) → c(x1) (4)
c(c(c(c(x1)))) → c(a(a(x1))) (5)
b(b(b(b(x1)))) → b(x1) (6)
b(c(c(c(x1)))) → b(a(a(x1))) (7)
a(b(b(b(x1)))) → a(x1) (8)
a(c(c(c(x1)))) → a(a(a(x1))) (9)
c(a(x1)) → c(a(c(b(x1)))) (10)
b(a(x1)) → b(a(c(b(x1)))) (11)
a(a(x1)) → a(a(c(b(x1)))) (12)

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
b1(b1(b1(b1(x1)))) → b1(x1) (13)
b1(b1(b1(b0(x1)))) → b0(x1) (14)
b1(b1(b1(b2(x1)))) → b2(x1) (15)
c1(b1(b1(b1(x1)))) → c1(x1) (16)
c1(b1(b1(b0(x1)))) → c0(x1) (17)
c1(b1(b1(b2(x1)))) → c2(x1) (18)
a1(b1(b1(b1(x1)))) → a1(x1) (19)
a1(b1(b1(b0(x1)))) → a0(x1) (20)
a1(b1(b1(b2(x1)))) → a2(x1) (21)
b0(c0(c0(c1(x1)))) → b2(a2(a1(x1))) (22)
b0(c0(c0(c0(x1)))) → b2(a2(a0(x1))) (23)
b0(c0(c0(c2(x1)))) → b2(a2(a2(x1))) (24)
c0(c0(c0(c1(x1)))) → c2(a2(a1(x1))) (25)
c0(c0(c0(c0(x1)))) → c2(a2(a0(x1))) (26)
c0(c0(c0(c2(x1)))) → c2(a2(a2(x1))) (27)
a0(c0(c0(c1(x1)))) → a2(a2(a1(x1))) (28)
a0(c0(c0(c0(x1)))) → a2(a2(a0(x1))) (29)
a0(c0(c0(c2(x1)))) → a2(a2(a2(x1))) (30)
b2(a1(x1)) → b2(a0(c1(b1(x1)))) (31)
b2(a0(x1)) → b2(a0(c1(b0(x1)))) (32)
b2(a2(x1)) → b2(a0(c1(b2(x1)))) (33)
c2(a1(x1)) → c2(a0(c1(b1(x1)))) (34)
c2(a0(x1)) → c2(a0(c1(b0(x1)))) (35)
c2(a2(x1)) → c2(a0(c1(b2(x1)))) (36)
a2(a1(x1)) → a2(a0(c1(b1(x1)))) (37)
a2(a0(x1)) → a2(a0(c1(b0(x1)))) (38)
a2(a2(x1)) → a2(a0(c1(b2(x1)))) (39)

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 +
4
[c1(x1)] = x1 +
0
[c2(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b1(x1)] = x1 +
4
[b2(x1)] = x1 +
0
[a0(x1)] = x1 +
0
[a1(x1)] = x1 +
5
[a2(x1)] = x1 +
1
all of the following rules can be deleted.
b1(b1(b1(b1(x1)))) → b1(x1) (13)
b1(b1(b1(b0(x1)))) → b0(x1) (14)
b1(b1(b1(b2(x1)))) → b2(x1) (15)
c1(b1(b1(b1(x1)))) → c1(x1) (16)
c1(b1(b1(b0(x1)))) → c0(x1) (17)
c1(b1(b1(b2(x1)))) → c2(x1) (18)
a1(b1(b1(b1(x1)))) → a1(x1) (19)
a1(b1(b1(b0(x1)))) → a0(x1) (20)
a1(b1(b1(b2(x1)))) → a2(x1) (21)
b0(c0(c0(c1(x1)))) → b2(a2(a1(x1))) (22)
b0(c0(c0(c0(x1)))) → b2(a2(a0(x1))) (23)
b0(c0(c0(c2(x1)))) → b2(a2(a2(x1))) (24)
c0(c0(c0(c1(x1)))) → c2(a2(a1(x1))) (25)
c0(c0(c0(c0(x1)))) → c2(a2(a0(x1))) (26)
c0(c0(c0(c2(x1)))) → c2(a2(a2(x1))) (27)
a0(c0(c0(c1(x1)))) → a2(a2(a1(x1))) (28)
a0(c0(c0(c0(x1)))) → a2(a2(a0(x1))) (29)
a0(c0(c0(c2(x1)))) → a2(a2(a2(x1))) (30)
b2(a1(x1)) → b2(a0(c1(b1(x1)))) (31)
b2(a2(x1)) → b2(a0(c1(b2(x1)))) (33)
c2(a1(x1)) → c2(a0(c1(b1(x1)))) (34)
c2(a2(x1)) → c2(a0(c1(b2(x1)))) (36)
a2(a1(x1)) → a2(a0(c1(b1(x1)))) (37)
a2(a2(x1)) → a2(a0(c1(b2(x1)))) (39)

1.1.1.1 R is empty

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