Certification Problem

Input (TPDB SRS_Standard/Bouchare_06/11)

The rewrite relation of the following TRS is considered.

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

{b(☐), a(☐)}

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

1.1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐)}

We obtain the transformed TRS
b(b(a(a(a(x1))))) → b(b(b(x1))) (10)
b(b(b(b(x1)))) → b(b(a(a(x1)))) (11)
b(b(a(a(x1)))) → b(b(a(b(a(x1))))) (12)
b(a(a(a(a(x1))))) → b(a(b(x1))) (13)
b(a(b(b(x1)))) → b(a(a(a(x1)))) (14)
b(a(a(a(x1)))) → b(a(a(b(a(x1))))) (15)
a(b(a(a(a(x1))))) → a(b(b(x1))) (16)
a(b(b(b(x1)))) → a(b(a(a(x1)))) (17)
a(b(a(a(x1)))) → a(b(a(b(a(x1))))) (18)
a(a(a(a(a(x1))))) → a(a(b(x1))) (19)
a(a(b(b(x1)))) → a(a(a(a(x1)))) (20)
a(a(a(a(x1)))) → a(a(a(b(a(x1))))) (21)

1.1.1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), a(☐)}

We obtain the transformed TRS
b(b(b(a(a(a(x1)))))) → b(b(b(b(x1)))) (22)
b(b(b(b(b(x1))))) → b(b(b(a(a(x1))))) (23)
b(b(b(a(a(x1))))) → b(b(b(a(b(a(x1)))))) (24)
b(b(a(a(a(a(x1)))))) → b(b(a(b(x1)))) (25)
b(b(a(b(b(x1))))) → b(b(a(a(a(x1))))) (26)
b(b(a(a(a(x1))))) → b(b(a(a(b(a(x1)))))) (27)
b(a(b(a(a(a(x1)))))) → b(a(b(b(x1)))) (28)
b(a(b(b(b(x1))))) → b(a(b(a(a(x1))))) (29)
b(a(b(a(a(x1))))) → b(a(b(a(b(a(x1)))))) (30)
b(a(a(a(a(a(x1)))))) → b(a(a(b(x1)))) (31)
b(a(a(b(b(x1))))) → b(a(a(a(a(x1))))) (32)
b(a(a(a(a(x1))))) → b(a(a(a(b(a(x1)))))) (33)
a(b(b(a(a(a(x1)))))) → a(b(b(b(x1)))) (34)
a(b(b(b(b(x1))))) → a(b(b(a(a(x1))))) (35)
a(b(b(a(a(x1))))) → a(b(b(a(b(a(x1)))))) (36)
a(b(a(a(a(a(x1)))))) → a(b(a(b(x1)))) (37)
a(b(a(b(b(x1))))) → a(b(a(a(a(x1))))) (38)
a(b(a(a(a(x1))))) → a(b(a(a(b(a(x1)))))) (39)
a(a(b(a(a(a(x1)))))) → a(a(b(b(x1)))) (40)
a(a(b(b(b(x1))))) → a(a(b(a(a(x1))))) (41)
a(a(b(a(a(x1))))) → a(a(b(a(b(a(x1)))))) (42)
a(a(a(a(a(a(x1)))))) → a(a(a(b(x1)))) (43)
a(a(a(b(b(x1))))) → a(a(a(a(a(x1))))) (44)
a(a(a(a(a(x1))))) → a(a(a(a(b(a(x1)))))) (45)

1.1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,...,7}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 8):

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

We obtain the labeled TRS

There are 192 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
[b0(x1)] = x1 +
237
[b4(x1)] = x1 +
189
[b2(x1)] = x1 +
189
[b6(x1)] = x1 +
190
[b1(x1)] = x1 +
189
[b5(x1)] = x1 +
189
[b3(x1)] = x1 +
190
[b7(x1)] = x1 +
190
[a0(x1)] = x1 +
189
[a4(x1)] = x1 +
211
[a2(x1)] = x1 +
0
[a6(x1)] = x1 +
1
[a1(x1)] = x1 +
190
[a5(x1)] = x1 +
190
[a3(x1)] = x1 +
190
[a7(x1)] = x1 +
194
all of the following rules can be deleted.

There are 192 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.