Certification Problem
Input (TPDB SRS_Relative/Zantema_06_relative/rel10)
The relative rewrite relation R/S is considered where R is the following TRS
|
b(p(b(x1))) |
→ |
b(q(b(x1))) |
(1) |
and S is the following TRS.
|
1(p(0(1(0(x1))))) |
→ |
p(x1) |
(2) |
|
q(x1) |
→ |
0(q(0(x1))) |
(3) |
|
q(x1) |
→ |
1(q(1(x1))) |
(4) |
|
q(x1) |
→ |
0(p(0(x1))) |
(5) |
|
q(x1) |
→ |
1(p(1(x1))) |
(6) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{1(☐), 0(☐), b(☐), q(☐), p(☐)}
We obtain the transformed TRS
|
1(b(p(b(x1)))) |
→ |
1(b(q(b(x1)))) |
(7) |
|
1(q(x1)) |
→ |
1(0(p(0(x1)))) |
(8) |
|
1(q(x1)) |
→ |
1(1(p(1(x1)))) |
(9) |
|
0(b(p(b(x1)))) |
→ |
0(b(q(b(x1)))) |
(10) |
|
0(q(x1)) |
→ |
0(0(p(0(x1)))) |
(11) |
|
0(q(x1)) |
→ |
0(1(p(1(x1)))) |
(12) |
|
b(b(p(b(x1)))) |
→ |
b(b(q(b(x1)))) |
(13) |
|
b(q(x1)) |
→ |
b(0(p(0(x1)))) |
(14) |
|
b(q(x1)) |
→ |
b(1(p(1(x1)))) |
(15) |
|
q(b(p(b(x1)))) |
→ |
q(b(q(b(x1)))) |
(16) |
|
q(q(x1)) |
→ |
q(0(p(0(x1)))) |
(17) |
|
q(q(x1)) |
→ |
q(1(p(1(x1)))) |
(18) |
|
p(b(p(b(x1)))) |
→ |
p(b(q(b(x1)))) |
(19) |
|
p(q(x1)) |
→ |
p(0(p(0(x1)))) |
(20) |
|
p(q(x1)) |
→ |
p(1(p(1(x1)))) |
(21) |
|
1(1(p(0(1(0(x1)))))) |
→ |
1(p(x1)) |
(22) |
|
1(q(x1)) |
→ |
1(0(q(0(x1)))) |
(23) |
|
1(q(x1)) |
→ |
1(1(q(1(x1)))) |
(24) |
|
0(1(p(0(1(0(x1)))))) |
→ |
0(p(x1)) |
(25) |
|
0(q(x1)) |
→ |
0(0(q(0(x1)))) |
(26) |
|
0(q(x1)) |
→ |
0(1(q(1(x1)))) |
(27) |
|
b(1(p(0(1(0(x1)))))) |
→ |
b(p(x1)) |
(28) |
|
b(q(x1)) |
→ |
b(0(q(0(x1)))) |
(29) |
|
b(q(x1)) |
→ |
b(1(q(1(x1)))) |
(30) |
|
q(1(p(0(1(0(x1)))))) |
→ |
q(p(x1)) |
(31) |
|
q(q(x1)) |
→ |
q(0(q(0(x1)))) |
(32) |
|
q(q(x1)) |
→ |
q(1(q(1(x1)))) |
(33) |
|
p(1(p(0(1(0(x1)))))) |
→ |
p(p(x1)) |
(34) |
|
p(q(x1)) |
→ |
p(0(q(0(x1)))) |
(35) |
|
p(q(x1)) |
→ |
p(1(q(1(x1)))) |
(36) |
1.1 Semantic Labeling
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,...,4}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 5):
| [1(x1)] |
= |
5x1 + 0 |
| [0(x1)] |
= |
5x1 + 1 |
| [b(x1)] |
= |
5x1 + 2 |
| [q(x1)] |
= |
5x1 + 3 |
| [p(x1)] |
= |
5x1 + 4 |
We obtain the labeled TRS
There are 150 ruless (increase limit for explicit display).
1.1.1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [10(x1)] |
= |
x1 +
|
| [11(x1)] |
= |
x1 +
|
| [12(x1)] |
= |
x1 +
|
| [13(x1)] |
= |
x1 +
|
| [14(x1)] |
= |
x1 +
|
| [00(x1)] |
= |
x1 +
|
| [01(x1)] |
= |
x1 +
|
| [02(x1)] |
= |
x1 +
|
| [03(x1)] |
= |
x1 +
|
| [04(x1)] |
= |
x1 +
|
| [b0(x1)] |
= |
x1 +
|
| [b1(x1)] |
= |
x1 +
|
| [b2(x1)] |
= |
x1 +
|
| [b3(x1)] |
= |
x1 +
|
| [b4(x1)] |
= |
x1 +
|
| [q0(x1)] |
= |
x1 +
|
| [q1(x1)] |
= |
x1 +
|
| [q2(x1)] |
= |
x1 +
|
| [q3(x1)] |
= |
x1 +
|
| [q4(x1)] |
= |
x1 +
|
| [p0(x1)] |
= |
x1 +
|
| [p1(x1)] |
= |
x1 +
|
| [p2(x1)] |
= |
x1 +
|
| [p3(x1)] |
= |
x1 +
|
| [p4(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
There are 134 ruless (increase limit for explicit display).
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.