Certification Problem

Input (TPDB SRS_Relative/Mixed_relative_SRS/dup03)

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

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

and S is the following TRS.

b(b(x1)) b(b(c(c(x1)))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Split

We split R in the relative problem D/R-D and R-D, where the rules D

a(a(a(a(x1)))) a(a(b(b(a(a(x1)))))) (2)
are deleted.

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(), b(), a()}

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

1.1.1 Closure Under Flat Contexts

Using the flat contexts

{c(), b(), a()}

We obtain the transformed TRS
c(c(a(a(a(a(x1)))))) c(c(a(a(b(b(a(a(x1)))))))) (13)
c(b(a(a(a(a(x1)))))) c(b(a(a(b(b(a(a(x1)))))))) (14)
c(a(a(a(a(a(x1)))))) c(a(a(a(b(b(a(a(x1)))))))) (15)
b(c(a(a(a(a(x1)))))) b(c(a(a(b(b(a(a(x1)))))))) (16)
b(b(a(a(a(a(x1)))))) b(b(a(a(b(b(a(a(x1)))))))) (17)
b(a(a(a(a(a(x1)))))) b(a(a(a(b(b(a(a(x1)))))))) (18)
a(c(a(a(a(a(x1)))))) a(c(a(a(b(b(a(a(x1)))))))) (19)
a(b(a(a(a(a(x1)))))) a(b(a(a(b(b(a(a(x1)))))))) (20)
a(a(a(a(a(a(x1)))))) a(a(a(a(b(b(a(a(x1)))))))) (21)
c(c(a(a(c(c(b(b(x1)))))))) c(c(b(b(a(a(b(b(a(a(x1)))))))))) (22)
c(c(b(b(x1)))) c(c(b(b(c(c(x1)))))) (23)
c(b(a(a(c(c(b(b(x1)))))))) c(b(b(b(a(a(b(b(a(a(x1)))))))))) (24)
c(b(b(b(x1)))) c(b(b(b(c(c(x1)))))) (25)
c(a(a(a(c(c(b(b(x1)))))))) c(a(b(b(a(a(b(b(a(a(x1)))))))))) (26)
c(a(b(b(x1)))) c(a(b(b(c(c(x1)))))) (27)
b(c(a(a(c(c(b(b(x1)))))))) b(c(b(b(a(a(b(b(a(a(x1)))))))))) (28)
b(c(b(b(x1)))) b(c(b(b(c(c(x1)))))) (29)
b(b(a(a(c(c(b(b(x1)))))))) b(b(b(b(a(a(b(b(a(a(x1)))))))))) (30)
b(b(b(b(x1)))) b(b(b(b(c(c(x1)))))) (31)
b(a(a(a(c(c(b(b(x1)))))))) b(a(b(b(a(a(b(b(a(a(x1)))))))))) (32)
b(a(b(b(x1)))) b(a(b(b(c(c(x1)))))) (33)
a(c(a(a(c(c(b(b(x1)))))))) a(c(b(b(a(a(b(b(a(a(x1)))))))))) (34)
a(c(b(b(x1)))) a(c(b(b(c(c(x1)))))) (35)
a(b(a(a(c(c(b(b(x1)))))))) a(b(b(b(a(a(b(b(a(a(x1)))))))))) (36)
a(b(b(b(x1)))) a(b(b(b(c(c(x1)))))) (37)
a(a(a(a(c(c(b(b(x1)))))))) a(a(b(b(a(a(b(b(a(a(x1)))))))))) (38)
a(a(b(b(x1)))) a(a(b(b(c(c(x1)))))) (39)

1.1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,8}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 9):

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

We obtain the labeled TRS

There are 243 ruless (increase limit for explicit display).

1.1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c0(x1)] = x1 +
0
[c3(x1)] = x1 +
0
[c6(x1)] = x1 +
0
[c1(x1)] = x1 +
0
[c4(x1)] = x1 +
1
[c7(x1)] = x1 +
1/2
[c2(x1)] = x1 +
0
[c5(x1)] = x1 +
0
[c8(x1)] = x1 +
0
[b0(x1)] = x1 +
0
[b3(x1)] = x1 +
1
[b6(x1)] = x1 +
4
[b1(x1)] = x1 +
0
[b4(x1)] = x1 +
1/2
[b7(x1)] = x1 +
0
[b2(x1)] = x1 +
4
[b5(x1)] = x1 +
1
[b8(x1)] = x1 +
0
[a0(x1)] = x1 +
4
[a3(x1)] = x1 +
0
[a6(x1)] = x1 +
1
[a1(x1)] = x1 +
0
[a4(x1)] = x1 +
0
[a7(x1)] = x1 +
1
[a2(x1)] = x1 +
4
[a5(x1)] = x1 +
0
[a8(x1)] = x1 +
1
all of the following rules can be deleted.

There are 204 ruless (increase limit for explicit display).

1.1.1.1.1.1 R is empty

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

1.2 Rule Removal

Using the matrix interpretations of dimension 4 with strict dimension 1 over the naturals
[c(x1)] =
1 0 0 0
0 0 7 1
0 0 0 0
0 0 4 0
· x1 +
0
0
0
0
[b(x1)] =
1 0 0 0
0 0 0 0
0 1 1 4
0 0 1 4
· x1 +
0
0
0
1
[a(x1)] =
1 1 0 0
0 2 0 0
0 0 0 1
0 1 0 0
· x1 +
0
0
0
0
all of the following rules can be deleted.
a(a(c(c(b(b(x1)))))) b(b(a(a(b(b(a(a(x1)))))))) (1)

1.2.1 R is empty

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