Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/25192)
The rewrite relation of the following TRS is considered.
0(x1) |
→ |
1(x1) |
(1) |
4(5(4(5(x1)))) |
→ |
4(4(5(5(x1)))) |
(2) |
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) |
→ |
2(x1) |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[5(x1)] |
= |
x1 +
|
[4(x1)] |
= |
x1 +
|
[2(x1)] |
= |
x1 +
|
[1(x1)] |
= |
x1 +
|
[0(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
0(x1) |
→ |
1(x1) |
(1) |
5(5(5(5(5(5(4(4(4(4(4(4(x1)))))))))))) |
→ |
2(x1) |
(3) |
1.1 Closure Under Flat Contexts
Using the flat contexts
{5(☐), 4(☐)}
We obtain the transformed TRS
5(4(5(4(5(x1))))) |
→ |
5(4(4(5(5(x1))))) |
(4) |
4(4(5(4(5(x1))))) |
→ |
4(4(4(5(5(x1))))) |
(5) |
1.1.1 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):
[5(x1)] |
= |
2x1 + 0 |
[4(x1)] |
= |
2x1 + 1 |
We obtain the labeled TRS
41(40(51(40(51(x1))))) |
→ |
41(41(40(50(51(x1))))) |
(6) |
41(40(51(40(50(x1))))) |
→ |
41(41(40(50(50(x1))))) |
(7) |
51(40(51(40(51(x1))))) |
→ |
51(41(40(50(51(x1))))) |
(8) |
51(40(51(40(50(x1))))) |
→ |
51(41(40(50(50(x1))))) |
(9) |
1.1.1.1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[50(x1)] |
= |
x1 +
|
[51(x1)] |
= |
x1 +
|
[40(x1)] |
= |
x1 +
|
[41(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
41(40(51(40(51(x1))))) |
→ |
41(41(40(50(51(x1))))) |
(6) |
41(40(51(40(50(x1))))) |
→ |
41(41(40(50(50(x1))))) |
(7) |
51(40(51(40(51(x1))))) |
→ |
51(41(40(50(51(x1))))) |
(8) |
51(40(51(40(50(x1))))) |
→ |
51(41(40(50(50(x1))))) |
(9) |
1.1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.