Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/212094)
The rewrite relation of the following TRS is considered.
|
0(1(1(2(x1)))) |
→ |
1(3(0(1(2(x1))))) |
(1) |
|
0(1(1(2(x1)))) |
→ |
1(3(1(0(2(x1))))) |
(2) |
|
0(1(1(2(x1)))) |
→ |
2(3(1(3(0(1(x1)))))) |
(3) |
|
0(1(2(0(x1)))) |
→ |
1(0(0(2(2(x1))))) |
(4) |
|
0(1(2(4(x1)))) |
→ |
0(2(3(4(1(x1))))) |
(5) |
|
0(1(2(4(x1)))) |
→ |
1(0(4(2(3(x1))))) |
(6) |
|
0(1(2(4(x1)))) |
→ |
1(2(3(0(4(x1))))) |
(7) |
|
0(1(2(4(x1)))) |
→ |
2(1(4(0(3(x1))))) |
(8) |
|
0(1(2(4(x1)))) |
→ |
1(2(3(4(0(5(x1)))))) |
(9) |
|
0(1(2(4(x1)))) |
→ |
2(1(1(0(3(4(x1)))))) |
(10) |
|
0(1(2(4(x1)))) |
→ |
4(1(2(2(3(0(x1)))))) |
(11) |
|
0(1(2(4(x1)))) |
→ |
4(5(0(2(3(1(x1)))))) |
(12) |
|
0(4(2(0(x1)))) |
→ |
0(4(2(3(0(x1))))) |
(13) |
|
0(4(2(0(x1)))) |
→ |
2(3(0(4(0(0(x1)))))) |
(14) |
|
0(4(2(0(x1)))) |
→ |
3(4(0(2(3(0(x1)))))) |
(15) |
|
0(4(2(0(x1)))) |
→ |
4(2(3(0(4(0(x1)))))) |
(16) |
|
0(4(2(0(x1)))) |
→ |
4(3(0(0(5(2(x1)))))) |
(17) |
|
0(4(2(4(x1)))) |
→ |
4(0(2(3(1(4(x1)))))) |
(18) |
|
0(4(2(4(x1)))) |
→ |
4(0(4(2(0(3(x1)))))) |
(19) |
|
5(1(2(0(x1)))) |
→ |
2(1(3(5(0(x1))))) |
(20) |
|
5(1(2(0(x1)))) |
→ |
3(1(0(2(5(x1))))) |
(21) |
|
5(1(2(0(x1)))) |
→ |
2(3(0(0(1(5(x1)))))) |
(22) |
|
5(1(2(4(x1)))) |
→ |
3(2(5(1(4(x1))))) |
(23) |
|
5(1(2(4(x1)))) |
→ |
5(2(1(3(4(x1))))) |
(24) |
|
5(1(2(4(x1)))) |
→ |
5(5(2(1(4(x1))))) |
(25) |
|
5(1(2(4(x1)))) |
→ |
0(3(4(5(2(1(x1)))))) |
(26) |
|
5(1(4(2(x1)))) |
→ |
3(4(5(2(1(x1))))) |
(27) |
|
5(1(4(2(x1)))) |
→ |
2(1(3(4(5(4(x1)))))) |
(28) |
|
5(1(4(2(x1)))) |
→ |
3(3(4(2(1(5(x1)))))) |
(29) |
|
5(4(1(4(x1)))) |
→ |
3(4(4(1(5(4(x1)))))) |
(30) |
|
5(4(1(4(x1)))) |
→ |
4(1(3(5(0(4(x1)))))) |
(31) |
|
5(4(1(4(x1)))) |
→ |
5(1(3(4(5(4(x1)))))) |
(32) |
|
5(4(1(4(x1)))) |
→ |
5(1(5(3(4(4(x1)))))) |
(33) |
|
5(4(2(0(x1)))) |
→ |
5(4(2(3(0(x1))))) |
(34) |
|
5(4(2(0(x1)))) |
→ |
0(1(2(3(4(5(x1)))))) |
(35) |
|
5(4(2(0(x1)))) |
→ |
4(5(3(3(0(2(x1)))))) |
(36) |
|
0(1(2(0(4(x1))))) |
→ |
0(2(4(1(3(0(x1)))))) |
(37) |
|
0(1(2(0(4(x1))))) |
→ |
2(0(3(4(0(1(x1)))))) |
(38) |
|
0(1(2(0(4(x1))))) |
→ |
4(0(2(3(1(0(x1)))))) |
(39) |
|
0(1(4(2(2(x1))))) |
→ |
1(2(3(4(0(2(x1)))))) |
(40) |
|
5(0(1(2(4(x1))))) |
→ |
3(0(2(1(4(5(x1)))))) |
(41) |
|
5(1(2(2(4(x1))))) |
→ |
5(2(2(1(4(5(x1)))))) |
(42) |
|
5(1(2(4(0(x1))))) |
→ |
5(0(2(1(3(4(x1)))))) |
(43) |
|
5(1(2(4(0(x1))))) |
→ |
5(0(3(2(1(4(x1)))))) |
(44) |
|
5(1(3(1(4(x1))))) |
→ |
1(5(1(3(4(0(x1)))))) |
(45) |
|
5(1(4(1(2(x1))))) |
→ |
2(1(5(3(4(1(x1)))))) |
(46) |
|
5(4(1(1(4(x1))))) |
→ |
1(1(3(4(5(4(x1)))))) |
(47) |
|
5(4(1(4(0(x1))))) |
→ |
3(4(0(4(5(1(x1)))))) |
(48) |
|
5(4(3(2(0(x1))))) |
→ |
5(4(0(3(2(3(x1)))))) |
(49) |
|
5(4(5(2(0(x1))))) |
→ |
5(0(5(4(4(2(x1)))))) |
(50) |
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 300 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 1800 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 1799 ruless (increase limit for explicit display).
1.1.1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
45(21(43(01(25(x1))))) |
→ |
35(02(25(43(01(45(21(x1))))))) |
(2151) |
1.1.1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
45#(21(43(01(25(x1))))) |
→ |
45#(21(x1)) |
(2152) |
1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [43(x1)] |
= |
x1 +
|
| [45(x1)] |
= |
x1 +
|
| [35(x1)] |
= |
x1 +
|
| [21(x1)] |
= |
x1 +
|
| [25(x1)] |
= |
x1 +
|
| [01(x1)] |
= |
x1 +
|
| [02(x1)] |
= |
x1 +
|
| [45#(x1)] |
= |
x1 +
|
together with the usable
rule
|
45(21(43(01(25(x1))))) |
→ |
35(02(25(43(01(45(21(x1))))))) |
(2151) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
45#(21(43(01(25(x1))))) |
→ |
45#(21(x1)) |
(2152) |
and
no rules
could be deleted.
1.1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.