Certification Problem
Input (TPDB SRS_Standard/ICFP_2010/213051)
The rewrite relation of the following TRS is considered.
0(0(1(x1))) |
→ |
0(1(2(0(3(x1))))) |
(1) |
0(0(1(x1))) |
→ |
0(1(2(0(3(2(x1)))))) |
(2) |
0(0(1(x1))) |
→ |
1(2(0(0(3(2(x1)))))) |
(3) |
0(1(0(x1))) |
→ |
2(0(3(1(2(0(x1)))))) |
(4) |
0(4(1(x1))) |
→ |
4(0(3(1(x1)))) |
(5) |
0(4(1(x1))) |
→ |
2(4(0(3(2(1(x1)))))) |
(6) |
0(4(1(x1))) |
→ |
2(4(0(5(3(1(x1)))))) |
(7) |
0(4(1(x1))) |
→ |
4(2(1(2(0(3(x1)))))) |
(8) |
4(1(0(x1))) |
→ |
1(2(0(4(x1)))) |
(9) |
0(0(1(1(x1)))) |
→ |
1(2(0(0(3(1(x1)))))) |
(10) |
0(0(2(1(x1)))) |
→ |
2(2(0(3(0(1(x1)))))) |
(11) |
0(0(4(1(x1)))) |
→ |
1(4(0(3(0(3(x1)))))) |
(12) |
0(0(5(1(x1)))) |
→ |
0(0(3(1(2(5(x1)))))) |
(13) |
0(1(1(0(x1)))) |
→ |
1(2(0(0(1(2(x1)))))) |
(14) |
0(1(4(1(x1)))) |
→ |
0(1(2(4(1(2(x1)))))) |
(15) |
0(1(5(0(x1)))) |
→ |
1(2(5(0(0(3(x1)))))) |
(16) |
0(2(4(1(x1)))) |
→ |
4(0(5(3(1(2(x1)))))) |
(17) |
0(4(1(0(x1)))) |
→ |
0(4(0(1(2(x1))))) |
(18) |
0(4(2(1(x1)))) |
→ |
4(0(3(2(2(1(x1)))))) |
(19) |
0(4(2(1(x1)))) |
→ |
4(0(3(5(1(2(x1)))))) |
(20) |
0(4(4(1(x1)))) |
→ |
2(4(0(3(4(1(x1)))))) |
(21) |
0(5(0(1(x1)))) |
→ |
0(2(0(3(5(1(x1)))))) |
(22) |
0(5(0(1(x1)))) |
→ |
5(2(0(3(0(1(x1)))))) |
(23) |
0(5(1(0(x1)))) |
→ |
0(1(2(0(3(5(x1)))))) |
(24) |
0(5(1(0(x1)))) |
→ |
2(0(1(3(5(0(x1)))))) |
(25) |
4(0(5(1(x1)))) |
→ |
1(4(0(5(3(2(x1)))))) |
(26) |
4(1(0(0(x1)))) |
→ |
0(4(1(2(0(x1))))) |
(27) |
4(1(0(1(x1)))) |
→ |
1(1(2(0(4(x1))))) |
(28) |
4(1(5(0(x1)))) |
→ |
1(2(0(5(4(x1))))) |
(29) |
4(1(5(0(x1)))) |
→ |
1(4(0(3(5(x1))))) |
(30) |
4(3(1(0(x1)))) |
→ |
1(4(2(0(3(x1))))) |
(31) |
4(3(1(0(x1)))) |
→ |
2(0(2(1(3(4(x1)))))) |
(32) |
4(3(1(0(x1)))) |
→ |
2(1(4(2(0(3(x1)))))) |
(33) |
4(3(1(0(x1)))) |
→ |
2(2(4(0(1(3(x1)))))) |
(34) |
5(0(1(0(x1)))) |
→ |
0(3(5(1(2(0(x1)))))) |
(35) |
5(0(1(0(x1)))) |
→ |
1(5(2(0(3(0(x1)))))) |
(36) |
5(4(1(0(x1)))) |
→ |
0(1(2(4(5(2(x1)))))) |
(37) |
0(0(5(5(1(x1))))) |
→ |
1(0(0(3(5(5(x1)))))) |
(38) |
0(1(0(5(0(x1))))) |
→ |
0(1(5(2(0(0(x1)))))) |
(39) |
0(2(5(0(1(x1))))) |
→ |
2(0(3(5(0(1(x1)))))) |
(40) |
0(3(1(0(0(x1))))) |
→ |
0(0(3(0(1(2(x1)))))) |
(41) |
0(4(1(4(1(x1))))) |
→ |
4(4(0(1(2(1(x1)))))) |
(42) |
0(5(5(4(1(x1))))) |
→ |
4(1(0(5(5(3(x1)))))) |
(43) |
4(1(5(0(0(x1))))) |
→ |
1(2(0(4(5(0(x1)))))) |
(44) |
4(1(5(5(0(x1))))) |
→ |
5(1(3(0(5(4(x1)))))) |
(45) |
4(3(1(0(1(x1))))) |
→ |
1(1(2(3(0(4(x1)))))) |
(46) |
4(3(1(1(0(x1))))) |
→ |
1(2(0(1(4(3(x1)))))) |
(47) |
4(3(1(5(0(x1))))) |
→ |
5(1(0(3(2(4(x1)))))) |
(48) |
4(4(1(0(5(x1))))) |
→ |
4(0(5(3(4(1(x1)))))) |
(49) |
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(0(0(x1))) |
→ |
3(0(2(1(0(x1))))) |
(50) |
1(0(0(x1))) |
→ |
2(3(0(2(1(0(x1)))))) |
(51) |
1(0(0(x1))) |
→ |
2(3(0(0(2(1(x1)))))) |
(52) |
0(1(0(x1))) |
→ |
0(2(1(3(0(2(x1)))))) |
(53) |
1(4(0(x1))) |
→ |
1(3(0(4(x1)))) |
(54) |
1(4(0(x1))) |
→ |
1(2(3(0(4(2(x1)))))) |
(55) |
1(4(0(x1))) |
→ |
1(3(5(0(4(2(x1)))))) |
(56) |
1(4(0(x1))) |
→ |
3(0(2(1(2(4(x1)))))) |
(57) |
0(1(4(x1))) |
→ |
4(0(2(1(x1)))) |
(58) |
1(1(0(0(x1)))) |
→ |
1(3(0(0(2(1(x1)))))) |
(59) |
1(2(0(0(x1)))) |
→ |
1(0(3(0(2(2(x1)))))) |
(60) |
1(4(0(0(x1)))) |
→ |
3(0(3(0(4(1(x1)))))) |
(61) |
1(5(0(0(x1)))) |
→ |
5(2(1(3(0(0(x1)))))) |
(62) |
0(1(1(0(x1)))) |
→ |
2(1(0(0(2(1(x1)))))) |
(63) |
1(4(1(0(x1)))) |
→ |
2(1(4(2(1(0(x1)))))) |
(64) |
0(5(1(0(x1)))) |
→ |
3(0(0(5(2(1(x1)))))) |
(65) |
1(4(2(0(x1)))) |
→ |
2(1(3(5(0(4(x1)))))) |
(66) |
0(1(4(0(x1)))) |
→ |
2(1(0(4(0(x1))))) |
(67) |
1(2(4(0(x1)))) |
→ |
1(2(2(3(0(4(x1)))))) |
(68) |
1(2(4(0(x1)))) |
→ |
2(1(5(3(0(4(x1)))))) |
(69) |
1(4(4(0(x1)))) |
→ |
1(4(3(0(4(2(x1)))))) |
(70) |
1(0(5(0(x1)))) |
→ |
1(5(3(0(2(0(x1)))))) |
(71) |
1(0(5(0(x1)))) |
→ |
1(0(3(0(2(5(x1)))))) |
(72) |
0(1(5(0(x1)))) |
→ |
5(3(0(2(1(0(x1)))))) |
(73) |
0(1(5(0(x1)))) |
→ |
0(5(3(1(0(2(x1)))))) |
(74) |
1(5(0(4(x1)))) |
→ |
2(3(5(0(4(1(x1)))))) |
(75) |
0(0(1(4(x1)))) |
→ |
0(2(1(4(0(x1))))) |
(76) |
1(0(1(4(x1)))) |
→ |
4(0(2(1(1(x1))))) |
(77) |
0(5(1(4(x1)))) |
→ |
4(5(0(2(1(x1))))) |
(78) |
0(5(1(4(x1)))) |
→ |
5(3(0(4(1(x1))))) |
(79) |
0(1(3(4(x1)))) |
→ |
3(0(2(4(1(x1))))) |
(80) |
0(1(3(4(x1)))) |
→ |
4(3(1(2(0(2(x1)))))) |
(81) |
0(1(3(4(x1)))) |
→ |
3(0(2(4(1(2(x1)))))) |
(82) |
0(1(3(4(x1)))) |
→ |
3(1(0(4(2(2(x1)))))) |
(83) |
0(1(0(5(x1)))) |
→ |
0(2(1(5(3(0(x1)))))) |
(84) |
0(1(0(5(x1)))) |
→ |
0(3(0(2(5(1(x1)))))) |
(85) |
0(1(4(5(x1)))) |
→ |
2(5(4(2(1(0(x1)))))) |
(86) |
1(5(5(0(0(x1))))) |
→ |
5(5(3(0(0(1(x1)))))) |
(87) |
0(5(0(1(0(x1))))) |
→ |
0(0(2(5(1(0(x1)))))) |
(88) |
1(0(5(2(0(x1))))) |
→ |
1(0(5(3(0(2(x1)))))) |
(89) |
0(0(1(3(0(x1))))) |
→ |
2(1(0(3(0(0(x1)))))) |
(90) |
1(4(1(4(0(x1))))) |
→ |
1(2(1(0(4(4(x1)))))) |
(91) |
1(4(5(5(0(x1))))) |
→ |
3(5(5(0(1(4(x1)))))) |
(92) |
0(0(5(1(4(x1))))) |
→ |
0(5(4(0(2(1(x1)))))) |
(93) |
0(5(5(1(4(x1))))) |
→ |
4(5(0(3(1(5(x1)))))) |
(94) |
1(0(1(3(4(x1))))) |
→ |
4(0(3(2(1(1(x1)))))) |
(95) |
0(1(1(3(4(x1))))) |
→ |
3(4(1(0(2(1(x1)))))) |
(96) |
0(5(1(3(4(x1))))) |
→ |
4(2(3(0(1(5(x1)))))) |
(97) |
5(0(1(4(4(x1))))) |
→ |
1(4(3(5(0(4(x1)))))) |
(98) |
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
There are 139 ruless (increase limit for explicit display).
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.