Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/212117)
The rewrite relation of the following TRS is considered.
|
0(1(1(x1))) |
→ |
0(1(2(3(4(1(x1)))))) |
(1) |
|
0(1(1(x1))) |
→ |
1(3(1(3(4(0(x1)))))) |
(2) |
|
0(1(1(x1))) |
→ |
5(1(3(0(3(1(x1)))))) |
(3) |
|
0(1(4(x1))) |
→ |
0(1(3(4(x1)))) |
(4) |
|
0(1(4(x1))) |
→ |
1(3(4(2(0(x1))))) |
(5) |
|
0(1(4(x1))) |
→ |
1(5(3(4(0(x1))))) |
(6) |
|
0(1(4(x1))) |
→ |
3(1(3(4(0(x1))))) |
(7) |
|
0(1(4(x1))) |
→ |
1(1(5(3(4(0(x1)))))) |
(8) |
|
0(1(4(x1))) |
→ |
1(2(3(4(3(0(x1)))))) |
(9) |
|
0(1(4(x1))) |
→ |
1(3(3(4(4(0(x1)))))) |
(10) |
|
0(1(4(x1))) |
→ |
1(3(4(5(0(5(x1)))))) |
(11) |
|
0(1(4(x1))) |
→ |
3(4(5(5(0(1(x1)))))) |
(12) |
|
1(2(4(x1))) |
→ |
1(2(3(4(1(x1))))) |
(13) |
|
1(2(4(x1))) |
→ |
5(3(4(1(2(x1))))) |
(14) |
|
1(2(4(x1))) |
→ |
1(5(2(3(4(3(x1)))))) |
(15) |
|
1(2(4(x1))) |
→ |
2(3(3(4(5(1(x1)))))) |
(16) |
|
5(2(1(x1))) |
→ |
1(2(2(3(5(4(x1)))))) |
(17) |
|
5(2(1(x1))) |
→ |
1(3(2(5(3(4(x1)))))) |
(18) |
|
5(2(4(x1))) |
→ |
0(5(2(3(4(x1))))) |
(19) |
|
5(2(4(x1))) |
→ |
5(5(3(4(2(x1))))) |
(20) |
|
5(2(4(x1))) |
→ |
0(3(4(4(5(2(x1)))))) |
(21) |
|
5(2(4(x1))) |
→ |
2(2(5(3(4(4(x1)))))) |
(22) |
|
5(2(4(x1))) |
→ |
2(3(3(4(5(5(x1)))))) |
(23) |
|
5(2(4(x1))) |
→ |
2(3(4(3(5(3(x1)))))) |
(24) |
|
5(2(4(x1))) |
→ |
2(5(3(4(0(5(x1)))))) |
(25) |
|
5(2(4(x1))) |
→ |
3(1(2(5(3(4(x1)))))) |
(26) |
|
5(2(4(x1))) |
→ |
5(2(3(4(0(3(x1)))))) |
(27) |
|
0(0(2(4(x1)))) |
→ |
0(4(3(0(2(x1))))) |
(28) |
|
0(1(1(5(x1)))) |
→ |
0(1(3(5(1(2(x1)))))) |
(29) |
|
0(1(2(4(x1)))) |
→ |
0(1(2(3(3(4(x1)))))) |
(30) |
|
0(1(4(5(x1)))) |
→ |
3(4(0(5(1(x1))))) |
(31) |
|
0(1(4(5(x1)))) |
→ |
3(4(5(3(0(1(x1)))))) |
(32) |
|
0(4(2(1(x1)))) |
→ |
0(4(1(2(3(4(x1)))))) |
(33) |
|
0(5(1(4(x1)))) |
→ |
0(0(3(5(4(1(x1)))))) |
(34) |
|
1(0(1(4(x1)))) |
→ |
3(4(0(1(2(1(x1)))))) |
(35) |
|
1(1(2(4(x1)))) |
→ |
1(3(4(1(2(x1))))) |
(36) |
|
1(2(2(4(x1)))) |
→ |
2(2(2(3(4(1(x1)))))) |
(37) |
|
1(2(4(2(x1)))) |
→ |
1(2(3(4(2(3(x1)))))) |
(38) |
|
1(5(2(1(x1)))) |
→ |
1(0(3(5(1(2(x1)))))) |
(39) |
|
1(5(2(4(x1)))) |
→ |
3(2(3(5(1(4(x1)))))) |
(40) |
|
1(5(2(4(x1)))) |
→ |
5(2(3(3(1(4(x1)))))) |
(41) |
|
5(0(1(4(x1)))) |
→ |
3(4(0(5(3(1(x1)))))) |
(42) |
|
5(2(5(1(x1)))) |
→ |
3(5(1(5(2(x1))))) |
(43) |
|
0(1(1(5(4(x1))))) |
→ |
0(0(5(4(1(1(x1)))))) |
(44) |
|
1(0(0(2(4(x1))))) |
→ |
0(2(0(4(4(1(x1)))))) |
(45) |
|
1(0(4(2(1(x1))))) |
→ |
1(2(3(4(1(0(x1)))))) |
(46) |
|
1(2(5(0(1(x1))))) |
→ |
2(3(5(1(0(1(x1)))))) |
(47) |
|
1(5(3(0(4(x1))))) |
→ |
5(1(3(4(0(4(x1)))))) |
(48) |
|
5(0(5(2(4(x1))))) |
→ |
4(0(0(5(5(2(x1)))))) |
(49) |
|
5(3(2(4(2(x1))))) |
→ |
2(1(5(3(4(2(x1)))))) |
(50) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
1(1(0(x1))) |
→ |
1(4(3(2(1(0(x1)))))) |
(51) |
|
1(1(0(x1))) |
→ |
0(4(3(1(3(1(x1)))))) |
(52) |
|
1(1(0(x1))) |
→ |
1(3(0(3(1(5(x1)))))) |
(53) |
|
4(1(0(x1))) |
→ |
4(3(1(0(x1)))) |
(54) |
|
4(1(0(x1))) |
→ |
0(2(4(3(1(x1))))) |
(55) |
|
4(1(0(x1))) |
→ |
0(4(3(5(1(x1))))) |
(56) |
|
4(1(0(x1))) |
→ |
0(4(3(1(3(x1))))) |
(57) |
|
4(1(0(x1))) |
→ |
0(4(3(5(1(1(x1)))))) |
(58) |
|
4(1(0(x1))) |
→ |
0(3(4(3(2(1(x1)))))) |
(59) |
|
4(1(0(x1))) |
→ |
0(4(4(3(3(1(x1)))))) |
(60) |
|
4(1(0(x1))) |
→ |
5(0(5(4(3(1(x1)))))) |
(61) |
|
4(1(0(x1))) |
→ |
1(0(5(5(4(3(x1)))))) |
(62) |
|
4(2(1(x1))) |
→ |
1(4(3(2(1(x1))))) |
(63) |
|
4(2(1(x1))) |
→ |
2(1(4(3(5(x1))))) |
(64) |
|
4(2(1(x1))) |
→ |
3(4(3(2(5(1(x1)))))) |
(65) |
|
4(2(1(x1))) |
→ |
1(5(4(3(3(2(x1)))))) |
(66) |
|
1(2(5(x1))) |
→ |
4(5(3(2(2(1(x1)))))) |
(67) |
|
1(2(5(x1))) |
→ |
4(3(5(2(3(1(x1)))))) |
(68) |
|
4(2(5(x1))) |
→ |
4(3(2(5(0(x1))))) |
(69) |
|
4(2(5(x1))) |
→ |
2(4(3(5(5(x1))))) |
(70) |
|
4(2(5(x1))) |
→ |
2(5(4(4(3(0(x1)))))) |
(71) |
|
4(2(5(x1))) |
→ |
4(4(3(5(2(2(x1)))))) |
(72) |
|
4(2(5(x1))) |
→ |
5(5(4(3(3(2(x1)))))) |
(73) |
|
4(2(5(x1))) |
→ |
3(5(3(4(3(2(x1)))))) |
(74) |
|
4(2(5(x1))) |
→ |
5(0(4(3(5(2(x1)))))) |
(75) |
|
4(2(5(x1))) |
→ |
4(3(5(2(1(3(x1)))))) |
(76) |
|
4(2(5(x1))) |
→ |
3(0(4(3(2(5(x1)))))) |
(77) |
|
4(2(0(0(x1)))) |
→ |
2(0(3(4(0(x1))))) |
(78) |
|
5(1(1(0(x1)))) |
→ |
2(1(5(3(1(0(x1)))))) |
(79) |
|
4(2(1(0(x1)))) |
→ |
4(3(3(2(1(0(x1)))))) |
(80) |
|
5(4(1(0(x1)))) |
→ |
1(5(0(4(3(x1))))) |
(81) |
|
5(4(1(0(x1)))) |
→ |
1(0(3(5(4(3(x1)))))) |
(82) |
|
1(2(4(0(x1)))) |
→ |
4(3(2(1(4(0(x1)))))) |
(83) |
|
4(1(5(0(x1)))) |
→ |
1(4(5(3(0(0(x1)))))) |
(84) |
|
4(1(0(1(x1)))) |
→ |
1(2(1(0(4(3(x1)))))) |
(85) |
|
4(2(1(1(x1)))) |
→ |
2(1(4(3(1(x1))))) |
(86) |
|
4(2(2(1(x1)))) |
→ |
1(4(3(2(2(2(x1)))))) |
(87) |
|
2(4(2(1(x1)))) |
→ |
3(2(4(3(2(1(x1)))))) |
(88) |
|
1(2(5(1(x1)))) |
→ |
2(1(5(3(0(1(x1)))))) |
(89) |
|
4(2(5(1(x1)))) |
→ |
4(1(5(3(2(3(x1)))))) |
(90) |
|
4(2(5(1(x1)))) |
→ |
4(1(3(3(2(5(x1)))))) |
(91) |
|
4(1(0(5(x1)))) |
→ |
1(3(5(0(4(3(x1)))))) |
(92) |
|
1(5(2(5(x1)))) |
→ |
2(5(1(5(3(x1))))) |
(93) |
|
4(5(1(1(0(x1))))) |
→ |
1(1(4(5(0(0(x1)))))) |
(94) |
|
4(2(0(0(1(x1))))) |
→ |
1(4(4(0(2(0(x1)))))) |
(95) |
|
1(2(4(0(1(x1))))) |
→ |
0(1(4(3(2(1(x1)))))) |
(96) |
|
1(0(5(2(1(x1))))) |
→ |
1(0(1(5(3(2(x1)))))) |
(97) |
|
4(0(3(5(1(x1))))) |
→ |
4(0(4(3(1(5(x1)))))) |
(98) |
|
4(2(5(0(5(x1))))) |
→ |
2(5(5(0(0(4(x1)))))) |
(99) |
|
2(4(2(3(5(x1))))) |
→ |
2(4(3(5(1(2(x1)))))) |
(100) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
There are 156 ruless (increase limit for explicit display).
1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.
-
The
1st
component contains the
pair
|
4#(2(5(0(5(x1))))) |
→ |
4#(x1) |
(251) |
1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [2(x1)] |
= |
1 · x1
|
| [5(x1)] |
= |
1 · x1
|
| [0(x1)] |
= |
1 · x1
|
| [4#(x1)] |
= |
1 · x1
|
having no usable rules (w.r.t. the implicit argument filter of the
reduction pair),
the
rule
could be deleted.
1.1.1.1.1 Size-Change Termination
Using size-change termination in combination with
the subterm criterion
one obtains the following initial size-change graphs.
|
4#(2(5(0(5(x1))))) |
→ |
4#(x1) |
(251) |
|
| 1 |
> |
1 |
As there is no critical graph in the transitive closure, there are no infinite chains.
-
The
2nd
component contains the
pair
|
1#(1(0(x1))) |
→ |
2#(1(0(x1))) |
(103) |
|
2#(4(2(3(5(x1))))) |
→ |
1#(2(x1)) |
(255) |
|
1#(1(0(x1))) |
→ |
1#(x1) |
(106) |
|
1#(1(0(x1))) |
→ |
1#(5(x1)) |
(108) |
|
1#(2(5(x1))) |
→ |
2#(2(1(x1))) |
(146) |
|
2#(4(2(3(5(x1))))) |
→ |
2#(x1) |
(256) |
|
1#(2(5(x1))) |
→ |
2#(1(x1)) |
(147) |
|
1#(2(5(x1))) |
→ |
1#(x1) |
(148) |
|
1#(2(4(0(x1)))) |
→ |
2#(1(4(0(x1)))) |
(192) |
|
1#(2(4(0(x1)))) |
→ |
1#(4(0(x1))) |
(193) |
|
1#(2(4(0(1(x1))))) |
→ |
2#(1(x1)) |
(239) |
|
1#(0(5(2(1(x1))))) |
→ |
2#(x1) |
(243) |
1.1.1.2 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [1(x1)] |
= |
1 · x1
|
| [0(x1)] |
= |
1 · x1
|
| [4(x1)] |
= |
1 · x1
|
| [3(x1)] |
= |
1 · x1
|
| [2(x1)] |
= |
1 · x1
|
| [5(x1)] |
= |
1 · x1
|
| [2#(x1)] |
= |
1 · x1
|
| [1#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
1(1(0(x1))) |
→ |
1(4(3(2(1(0(x1)))))) |
(51) |
|
1(1(0(x1))) |
→ |
0(4(3(1(3(1(x1)))))) |
(52) |
|
1(1(0(x1))) |
→ |
1(3(0(3(1(5(x1)))))) |
(53) |
|
1(2(5(x1))) |
→ |
4(5(3(2(2(1(x1)))))) |
(67) |
|
1(2(5(x1))) |
→ |
4(3(5(2(3(1(x1)))))) |
(68) |
|
1(2(4(0(x1)))) |
→ |
4(3(2(1(4(0(x1)))))) |
(83) |
|
1(2(5(1(x1)))) |
→ |
2(1(5(3(0(1(x1)))))) |
(89) |
|
1(5(2(5(x1)))) |
→ |
2(5(1(5(3(x1))))) |
(93) |
|
1(2(4(0(1(x1))))) |
→ |
0(1(4(3(2(1(x1)))))) |
(96) |
|
1(0(5(2(1(x1))))) |
→ |
1(0(1(5(3(2(x1)))))) |
(97) |
|
2(4(2(1(x1)))) |
→ |
3(2(4(3(2(1(x1)))))) |
(88) |
|
2(4(2(3(5(x1))))) |
→ |
2(4(3(5(1(2(x1)))))) |
(100) |
|
5(1(1(0(x1)))) |
→ |
2(1(5(3(1(0(x1)))))) |
(79) |
|
5(4(1(0(x1)))) |
→ |
1(5(0(4(3(x1))))) |
(81) |
|
5(4(1(0(x1)))) |
→ |
1(0(3(5(4(3(x1)))))) |
(82) |
|
4(0(3(5(1(x1))))) |
→ |
4(0(4(3(1(5(x1)))))) |
(98) |
(w.r.t. the implicit argument filter of the reduction pair),
the
rule
could be deleted.
1.1.1.2.1 Dependency Graph Processor
The dependency pairs are split into 1
component.