Certification Problem
Input (TPDB SRS_Standard/Bouchare_06/04)
The rewrite relation of the following TRS is considered.
|
b(b(b(x1))) |
→ |
a(b(x1)) |
(1) |
|
a(a(x1)) |
→ |
a(b(a(x1))) |
(2) |
|
a(a(a(x1))) |
→ |
a(b(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
{b(☐), a(☐)}
We obtain the transformed TRS
|
b(b(b(b(x1)))) |
→ |
b(a(b(x1))) |
(4) |
|
b(a(a(x1))) |
→ |
b(a(b(a(x1)))) |
(5) |
|
b(a(a(a(x1)))) |
→ |
b(a(b(b(x1)))) |
(6) |
|
a(b(b(b(x1)))) |
→ |
a(a(b(x1))) |
(7) |
|
a(a(a(x1))) |
→ |
a(a(b(a(x1)))) |
(8) |
|
a(a(a(a(x1)))) |
→ |
a(a(b(b(x1)))) |
(9) |
1.1 Semantic Labeling
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,1}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 2):
| [b(x1)] |
= |
2x1 + 0 |
| [a(x1)] |
= |
2x1 + 1 |
We obtain the labeled TRS
|
b0(b0(b0(b0(x1)))) |
→ |
b1(a0(b0(x1))) |
(10) |
|
b0(b0(b0(b1(x1)))) |
→ |
b1(a0(b1(x1))) |
(11) |
|
a0(b0(b0(b0(x1)))) |
→ |
a1(a0(b0(x1))) |
(12) |
|
a0(b0(b0(b1(x1)))) |
→ |
a1(a0(b1(x1))) |
(13) |
|
b1(a1(a0(x1))) |
→ |
b1(a0(b1(a0(x1)))) |
(14) |
|
b1(a1(a1(x1))) |
→ |
b1(a0(b1(a1(x1)))) |
(15) |
|
a1(a1(a0(x1))) |
→ |
a1(a0(b1(a0(x1)))) |
(16) |
|
a1(a1(a1(x1))) |
→ |
a1(a0(b1(a1(x1)))) |
(17) |
|
b1(a1(a1(a0(x1)))) |
→ |
b1(a0(b0(b0(x1)))) |
(18) |
|
b1(a1(a1(a1(x1)))) |
→ |
b1(a0(b0(b1(x1)))) |
(19) |
|
a1(a1(a1(a0(x1)))) |
→ |
a1(a0(b0(b0(x1)))) |
(20) |
|
a1(a1(a1(a1(x1)))) |
→ |
a1(a0(b0(b1(x1)))) |
(21) |
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 +
|
| [b1(x1)] |
= |
x1 +
|
| [a0(x1)] |
= |
x1 +
|
| [a1(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
|
b0(b0(b0(b0(x1)))) |
→ |
b1(a0(b0(x1))) |
(10) |
|
b0(b0(b0(b1(x1)))) |
→ |
b1(a0(b1(x1))) |
(11) |
|
a0(b0(b0(b0(x1)))) |
→ |
a1(a0(b0(x1))) |
(12) |
|
a0(b0(b0(b1(x1)))) |
→ |
a1(a0(b1(x1))) |
(13) |
|
b1(a1(a0(x1))) |
→ |
b1(a0(b1(a0(x1)))) |
(14) |
|
b1(a1(a1(x1))) |
→ |
b1(a0(b1(a1(x1)))) |
(15) |
|
a1(a1(a0(x1))) |
→ |
a1(a0(b1(a0(x1)))) |
(16) |
|
a1(a1(a1(x1))) |
→ |
a1(a0(b1(a1(x1)))) |
(17) |
|
b1(a1(a1(a0(x1)))) |
→ |
b1(a0(b0(b0(x1)))) |
(18) |
|
b1(a1(a1(a1(x1)))) |
→ |
b1(a0(b0(b1(x1)))) |
(19) |
|
a1(a1(a1(a0(x1)))) |
→ |
a1(a0(b0(b0(x1)))) |
(20) |
|
a1(a1(a1(a1(x1)))) |
→ |
a1(a0(b0(b1(x1)))) |
(21) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.