Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/157388)
The rewrite relation of the following TRS is considered.
There are 151 ruless (increase limit for explicit display).
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
There are 290 ruless (increase limit for explicit display).
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
0#(1(0(1(x1)))) |
→ |
0#(0(x1)) |
(392) |
|
4#(5(5(1(1(x1))))) |
→ |
4#(x1) |
(393) |
|
4#(5(3(0(1(x1))))) |
→ |
0#(x1) |
(169) |
|
4#(5(3(0(1(x1))))) |
→ |
0#(x1) |
(169) |
|
0#(4(5(1(1(x1))))) |
→ |
0#(1(x1)) |
(375) |
|
4#(4(3(0(1(x1))))) |
→ |
4#(x1) |
(356) |
|
0#(1(0(1(x1)))) |
→ |
0#(x1) |
(349) |
|
4#(3(0(1(1(x1))))) |
→ |
0#(x1) |
(344) |
|
0#(4(3(0(1(x1))))) |
→ |
4#(0(x1)) |
(340) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
4#(3(5(0(1(x1))))) |
→ |
4#(x1) |
(280) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
4#(3(5(0(1(x1))))) |
→ |
4#(x1) |
(280) |
|
0#(4(1(0(1(x1))))) |
→ |
0#(4(0(1(x1)))) |
(270) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
0#(4(3(0(1(x1))))) |
→ |
0#(x1) |
(267) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
0#(4(3(1(1(x1))))) |
→ |
0#(x1) |
(257) |
|
0#(0(3(1(1(x1))))) |
→ |
0#(x1) |
(238) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(4(1(x1))) |
(235) |
|
0#(4(1(0(1(x1))))) |
→ |
4#(0(1(x1))) |
(226) |
|
0#(4(3(0(1(x1))))) |
→ |
0#(4(0(x1))) |
(225) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(0(x1)) |
(224) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(5(x1)) |
(218) |
|
0#(0(3(1(1(x1))))) |
→ |
0#(0(x1)) |
(212) |
|
0#(4(3(0(1(x1))))) |
→ |
4#(x1) |
(210) |
|
4#(3(0(1(1(x1))))) |
→ |
4#(x1) |
(209) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
0#(4(5(1(1(x1))))) |
→ |
4#(0(1(x1))) |
(189) |
|
4#(3(0(1(x1)))) |
→ |
4#(3(x1)) |
(177) |
|
4#(5(3(0(1(x1))))) |
→ |
0#(x1) |
(169) |
|
4#(5(3(0(1(x1))))) |
→ |
0#(x1) |
(169) |
|
4#(5(3(0(1(x1))))) |
→ |
0#(x1) |
(169) |
1.1.1 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [0#(x1)] |
=
|
x1 + 1 |
| [1(x1)] |
=
|
x1 + 0 |
| [4(x1)] |
=
|
x1 + 0 |
| [5(x1)] |
=
|
x1 + 0 |
| [3(x1)] |
=
|
x1 + 0 |
| [4#(x1)] |
=
|
x1 + 0 |
| [0(x1)] |
=
|
x1 + 1 |
| [2(x1)] |
=
|
x1 + 0 |
together with the usable
rulesThere are 151 ruless (increase limit for explicit display).
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
4#(4(3(0(1(x1))))) |
→ |
4#(x1) |
(356) |
|
0#(1(0(1(x1)))) |
→ |
0#(x1) |
(349) |
|
0#(4(3(0(1(x1))))) |
→ |
4#(0(x1)) |
(340) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
4#(3(5(0(1(x1))))) |
→ |
4#(x1) |
(280) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
4#(3(5(0(1(x1))))) |
→ |
4#(x1) |
(280) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
0#(4(3(0(1(x1))))) |
→ |
0#(x1) |
(267) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
0#(0(3(1(1(x1))))) |
→ |
0#(x1) |
(238) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(4(1(x1))) |
(235) |
|
0#(4(1(0(1(x1))))) |
→ |
4#(0(1(x1))) |
(226) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(5(x1)) |
(218) |
|
0#(4(3(0(1(x1))))) |
→ |
4#(x1) |
(210) |
|
4#(3(0(1(1(x1))))) |
→ |
4#(x1) |
(209) |
|
4#(3(0(1(x1)))) |
→ |
4#(x1) |
(203) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
4#(5(3(0(1(x1))))) |
→ |
4#(x1) |
(192) |
|
4#(3(0(1(x1)))) |
→ |
4#(3(x1)) |
(177) |
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
2nd
component contains the
pair
|
4#(1(0(1(x1)))) |
→ |
4#(1(x1)) |
(362) |
1.1.2 Reduction Pair Processor with Usable Rules
Using the Max-polynomial interpretation
| [0#(x1)] |
=
|
2 |
| [1(x1)] |
=
|
x1 + 0 |
| [4(x1)] |
=
|
x1 + 0 |
| [5(x1)] |
=
|
x1 + 0 |
| [3(x1)] |
=
|
x1 + 0 |
| [4#(x1)] |
=
|
x1 + 0 |
| [0(x1)] |
=
|
x1 + 1 |
| [2(x1)] |
=
|
x1 + 0 |
together with the usable
rulesThere are 151 ruless (increase limit for explicit display).
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
4#(1(0(1(x1)))) |
→ |
4#(1(x1)) |
(362) |
could be deleted.
1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 0
components.