Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/3336)
The rewrite relation of the following TRS is considered.
|
2(5(x1)) |
→ |
1(3(3(0(1(0(x1)))))) |
(1) |
|
2(5(x1)) |
→ |
2(2(0(5(0(1(x1)))))) |
(2) |
|
3(5(x1)) |
→ |
1(3(2(0(0(1(x1)))))) |
(3) |
|
3(5(x1)) |
→ |
3(2(0(5(3(0(x1)))))) |
(4) |
|
4(5(x1)) |
→ |
2(2(1(3(2(1(x1)))))) |
(5) |
|
4(5(x1)) |
→ |
3(2(0(5(0(0(x1)))))) |
(6) |
|
1(2(5(x1))) |
→ |
1(0(5(0(5(4(x1)))))) |
(7) |
|
1(2(5(x1))) |
→ |
1(2(2(1(0(1(x1)))))) |
(8) |
|
1(2(5(x1))) |
→ |
2(0(1(3(1(0(x1)))))) |
(9) |
|
1(4(5(x1))) |
→ |
1(2(4(0(2(1(x1)))))) |
(10) |
|
2(5(1(x1))) |
→ |
2(2(2(1(2(3(x1)))))) |
(11) |
|
2(5(2(x1))) |
→ |
4(0(2(2(3(3(x1)))))) |
(12) |
|
2(5(3(x1))) |
→ |
2(0(4(1(3(3(x1)))))) |
(13) |
|
2(5(4(x1))) |
→ |
2(0(5(1(0(1(x1)))))) |
(14) |
|
3(2(5(x1))) |
→ |
3(2(0(1(0(5(x1)))))) |
(15) |
|
3(4(2(x1))) |
→ |
3(4(0(2(2(2(x1)))))) |
(16) |
|
3(5(1(x1))) |
→ |
0(4(2(0(0(5(x1)))))) |
(17) |
|
3(5(1(x1))) |
→ |
0(4(2(2(3(4(x1)))))) |
(18) |
|
3(5(1(x1))) |
→ |
2(1(4(1(0(1(x1)))))) |
(19) |
|
3(5(2(x1))) |
→ |
0(4(3(2(2(2(x1)))))) |
(20) |
|
3(5(2(x1))) |
→ |
2(0(2(2(3(0(x1)))))) |
(21) |
|
3(5(2(x1))) |
→ |
2(3(3(2(1(2(x1)))))) |
(22) |
|
3(5(3(x1))) |
→ |
0(2(4(3(3(0(x1)))))) |
(23) |
|
3(5(3(x1))) |
→ |
0(5(4(3(3(0(x1)))))) |
(24) |
|
3(5(3(x1))) |
→ |
2(3(4(0(4(2(x1)))))) |
(25) |
|
3(5(4(x1))) |
→ |
0(2(0(5(0(0(x1)))))) |
(26) |
|
3(5(4(x1))) |
→ |
0(5(0(0(1(2(x1)))))) |
(27) |
|
3(5(5(x1))) |
→ |
0(5(4(1(0(5(x1)))))) |
(28) |
|
4(5(1(x1))) |
→ |
2(1(0(5(3(3(x1)))))) |
(29) |
|
4(5(2(x1))) |
→ |
0(5(1(0(0(4(x1)))))) |
(30) |
|
4(5(4(x1))) |
→ |
2(2(1(0(4(2(x1)))))) |
(31) |
|
4(5(4(x1))) |
→ |
3(2(0(3(2(0(x1)))))) |
(32) |
|
5(5(3(x1))) |
→ |
5(1(0(1(2(2(x1)))))) |
(33) |
|
5(5(4(x1))) |
→ |
5(1(0(4(2(2(x1)))))) |
(34) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Closure Under Flat Contexts
Using the flat contexts
{5(☐), 4(☐), 3(☐), 2(☐), 1(☐), 0(☐)}
We obtain the transformed TRS
There are 204 ruless (increase limit for explicit display).
1.1 Semantic Labeling
The following interpretations form a
model
of the rules.
As carrier we take the set
{0,...,5}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 6):
| [5(x1)] |
= |
6x1 + 0 |
| [4(x1)] |
= |
6x1 + 1 |
| [3(x1)] |
= |
6x1 + 2 |
| [2(x1)] |
= |
6x1 + 3 |
| [1(x1)] |
= |
6x1 + 4 |
| [0(x1)] |
= |
6x1 + 5 |
We obtain the labeled TRS
There are 1224 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
| [50(x1)] |
= |
x1 +
|
| [51(x1)] |
= |
x1 +
|
| [52(x1)] |
= |
x1 +
|
| [53(x1)] |
= |
x1 +
|
| [54(x1)] |
= |
x1 +
|
| [55(x1)] |
= |
x1 +
|
| [40(x1)] |
= |
x1 +
|
| [41(x1)] |
= |
x1 +
|
| [42(x1)] |
= |
x1 +
|
| [43(x1)] |
= |
x1 +
|
| [44(x1)] |
= |
x1 +
|
| [45(x1)] |
= |
x1 +
|
| [30(x1)] |
= |
x1 +
|
| [31(x1)] |
= |
x1 +
|
| [32(x1)] |
= |
x1 +
|
| [33(x1)] |
= |
x1 +
|
| [34(x1)] |
= |
x1 +
|
| [35(x1)] |
= |
x1 +
|
| [20(x1)] |
= |
x1 +
|
| [21(x1)] |
= |
x1 +
|
| [22(x1)] |
= |
x1 +
|
| [23(x1)] |
= |
x1 +
|
| [24(x1)] |
= |
x1 +
|
| [25(x1)] |
= |
x1 +
|
| [10(x1)] |
= |
x1 +
|
| [11(x1)] |
= |
x1 +
|
| [12(x1)] |
= |
x1 +
|
| [13(x1)] |
= |
x1 +
|
| [14(x1)] |
= |
x1 +
|
| [15(x1)] |
= |
x1 +
|
| [00(x1)] |
= |
x1 +
|
| [01(x1)] |
= |
x1 +
|
| [02(x1)] |
= |
x1 +
|
| [03(x1)] |
= |
x1 +
|
| [04(x1)] |
= |
x1 +
|
| [05(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
There are 1224 ruless (increase limit for explicit display).
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.