Certification Problem
Input (TPDB SRS_Relative/Waldmann_06_relative/r6)
The relative rewrite relation R/S is considered where R is the following TRS
b(b(x1)) |
→ |
c(b(c(x1))) |
(1) |
c(c(c(x1))) |
→ |
a(x1) |
(2) |
and S is the following TRS.
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 3 with strict dimension 1 over the naturals
[c(x1)] |
= |
· x1 +
|
[b(x1)] |
= |
· x1 +
|
[a(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
b(b(x1)) |
→ |
c(b(c(x1))) |
(1) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{c(☐), b(☐), a(☐)}
We obtain the transformed TRS
c(c(c(c(x1)))) |
→ |
c(a(x1)) |
(4) |
b(c(c(c(x1)))) |
→ |
b(a(x1)) |
(5) |
a(c(c(c(x1)))) |
→ |
a(a(x1)) |
(6) |
c(a(x1)) |
→ |
c(a(c(b(x1)))) |
(7) |
b(a(x1)) |
→ |
b(a(c(b(x1)))) |
(8) |
a(a(x1)) |
→ |
a(a(c(b(x1)))) |
(9) |
1.1.1 Semantic Labeling
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,1,2}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 3):
[c(x1)] |
= |
3x1 + 0 |
[b(x1)] |
= |
3x1 + 1 |
[a(x1)] |
= |
3x1 + 2 |
We obtain the labeled TRS
c0(c0(c0(c0(x1)))) |
→ |
c2(a0(x1)) |
(10) |
c0(c0(c0(c2(x1)))) |
→ |
c2(a2(x1)) |
(11) |
c0(c0(c0(c1(x1)))) |
→ |
c2(a1(x1)) |
(12) |
a0(c0(c0(c0(x1)))) |
→ |
a2(a0(x1)) |
(13) |
a0(c0(c0(c2(x1)))) |
→ |
a2(a2(x1)) |
(14) |
a0(c0(c0(c1(x1)))) |
→ |
a2(a1(x1)) |
(15) |
b0(c0(c0(c0(x1)))) |
→ |
b2(a0(x1)) |
(16) |
b0(c0(c0(c2(x1)))) |
→ |
b2(a2(x1)) |
(17) |
b0(c0(c0(c1(x1)))) |
→ |
b2(a1(x1)) |
(18) |
c2(a0(x1)) |
→ |
c2(a0(c1(b0(x1)))) |
(19) |
c2(a2(x1)) |
→ |
c2(a0(c1(b2(x1)))) |
(20) |
c2(a1(x1)) |
→ |
c2(a0(c1(b1(x1)))) |
(21) |
a2(a0(x1)) |
→ |
a2(a0(c1(b0(x1)))) |
(22) |
a2(a2(x1)) |
→ |
a2(a0(c1(b2(x1)))) |
(23) |
a2(a1(x1)) |
→ |
a2(a0(c1(b1(x1)))) |
(24) |
b2(a0(x1)) |
→ |
b2(a0(c1(b0(x1)))) |
(25) |
b2(a2(x1)) |
→ |
b2(a0(c1(b2(x1)))) |
(26) |
b2(a1(x1)) |
→ |
b2(a0(c1(b1(x1)))) |
(27) |
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 +
|
[c1(x1)] |
= |
x1 +
|
[c2(x1)] |
= |
x1 +
|
[b0(x1)] |
= |
x1 +
|
[b1(x1)] |
= |
x1 +
|
[b2(x1)] |
= |
x1 +
|
[a0(x1)] |
= |
x1 +
|
[a1(x1)] |
= |
x1 +
|
[a2(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
c0(c0(c0(c0(x1)))) |
→ |
c2(a0(x1)) |
(10) |
c0(c0(c0(c2(x1)))) |
→ |
c2(a2(x1)) |
(11) |
c0(c0(c0(c1(x1)))) |
→ |
c2(a1(x1)) |
(12) |
a0(c0(c0(c0(x1)))) |
→ |
a2(a0(x1)) |
(13) |
a0(c0(c0(c2(x1)))) |
→ |
a2(a2(x1)) |
(14) |
a0(c0(c0(c1(x1)))) |
→ |
a2(a1(x1)) |
(15) |
b0(c0(c0(c0(x1)))) |
→ |
b2(a0(x1)) |
(16) |
b0(c0(c0(c2(x1)))) |
→ |
b2(a2(x1)) |
(17) |
b0(c0(c0(c1(x1)))) |
→ |
b2(a1(x1)) |
(18) |
c2(a2(x1)) |
→ |
c2(a0(c1(b2(x1)))) |
(20) |
c2(a1(x1)) |
→ |
c2(a0(c1(b1(x1)))) |
(21) |
a2(a2(x1)) |
→ |
a2(a0(c1(b2(x1)))) |
(23) |
a2(a1(x1)) |
→ |
a2(a0(c1(b1(x1)))) |
(24) |
b2(a2(x1)) |
→ |
b2(a0(c1(b2(x1)))) |
(26) |
b2(a1(x1)) |
→ |
b2(a0(c1(b1(x1)))) |
(27) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.