Certification Problem

Input (TPDB SRS_Standard/Mixed_SRS/3)

The rewrite relation of the following TRS is considered.

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

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(a(a(x1))) → c(a(b(a(x1)))) (3)
c(b(a(b(x1)))) → c(a(c(a(x1)))) (4)
b(a(a(x1))) → b(a(b(a(x1)))) (5)
b(b(a(b(x1)))) → b(a(c(a(x1)))) (6)
a(a(a(x1))) → a(a(b(a(x1)))) (7)
a(b(a(b(x1)))) → a(a(c(a(x1)))) (8)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(c(a(a(x1)))) → c(c(a(b(a(x1))))) (9)
c(c(b(a(b(x1))))) → c(c(a(c(a(x1))))) (10)
c(b(a(a(x1)))) → c(b(a(b(a(x1))))) (11)
c(b(b(a(b(x1))))) → c(b(a(c(a(x1))))) (12)
c(a(a(a(x1)))) → c(a(a(b(a(x1))))) (13)
c(a(b(a(b(x1))))) → c(a(a(c(a(x1))))) (14)
b(c(a(a(x1)))) → b(c(a(b(a(x1))))) (15)
b(c(b(a(b(x1))))) → b(c(a(c(a(x1))))) (16)
b(b(a(a(x1)))) → b(b(a(b(a(x1))))) (17)
b(b(b(a(b(x1))))) → b(b(a(c(a(x1))))) (18)
b(a(a(a(x1)))) → b(a(a(b(a(x1))))) (19)
b(a(b(a(b(x1))))) → b(a(a(c(a(x1))))) (20)
a(c(a(a(x1)))) → a(c(a(b(a(x1))))) (21)
a(c(b(a(b(x1))))) → a(c(a(c(a(x1))))) (22)
a(b(a(a(x1)))) → a(b(a(b(a(x1))))) (23)
a(b(b(a(b(x1))))) → a(b(a(c(a(x1))))) (24)
a(a(a(a(x1)))) → a(a(a(b(a(x1))))) (25)
a(a(b(a(b(x1))))) → a(a(a(c(a(x1))))) (26)

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 162 ruless (increase limit for explicit display).

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c3(x1)] = x1 +
1
[c6(x1)] = x1 +
0
[c4(x1)] = x1 +
1
[c7(x1)] = x1 +
1/2
[c2(x1)] = x1 +
0
[c5(x1)] = x1 +
0
[c8(x1)] = x1 +
3
[b0(x1)] = x1 +
1
[b3(x1)] = x1 +
1
[b6(x1)] = x1 +
1/2
[b1(x1)] = x1 +
1
[b4(x1)] = x1 +
1
[b7(x1)] = x1 +
1
[b2(x1)] = x1 +
0
[b5(x1)] = x1 +
13
[b8(x1)] = x1 +
14
[a0(x1)] = x1 +
0
[a3(x1)] = x1 +
1/2
[a6(x1)] = x1 +
0
[a1(x1)] = x1 +
1/2
[a4(x1)] = x1 +
1
[a7(x1)] = x1 +
0
[a2(x1)] = x1 +
3
[a5(x1)] = x1 +
14
[a8(x1)] = x1 +
16
all of the following rules can be deleted.

There are 162 ruless (increase limit for explicit display).

1.1.1.1.1 R is empty

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