Certification Problem
Input (COPS 950)
We consider the TRS containing the following rules:
|
0(0(1(0(x)))) |
→ |
0(2(0(0(3(1(x)))))) |
(1) |
|
0(0(1(0(x)))) |
→ |
0(2(0(4(1(0(x)))))) |
(2) |
|
0(0(1(0(x)))) |
→ |
2(0(0(0(2(1(x)))))) |
(3) |
|
3(0(1(0(x)))) |
→ |
0(2(3(1(0(x))))) |
(4) |
|
3(0(1(0(x)))) |
→ |
3(1(0(0(2(x))))) |
(5) |
|
3(0(1(0(x)))) |
→ |
3(1(1(0(0(x))))) |
(6) |
|
3(0(1(0(x)))) |
→ |
3(1(2(0(0(x))))) |
(7) |
|
3(0(1(0(x)))) |
→ |
3(1(5(0(0(x))))) |
(8) |
|
3(0(1(0(x)))) |
→ |
3(5(1(0(0(x))))) |
(9) |
|
3(0(1(0(x)))) |
→ |
5(0(3(1(0(x))))) |
(10) |
|
3(0(1(0(x)))) |
→ |
2(0(2(3(1(0(x)))))) |
(11) |
|
3(0(1(0(x)))) |
→ |
2(2(0(3(1(0(x)))))) |
(12) |
|
3(0(1(0(x)))) |
→ |
3(1(5(0(0(0(x)))))) |
(13) |
|
3(0(1(0(x)))) |
→ |
3(1(5(0(2(0(x)))))) |
(14) |
|
3(0(1(0(x)))) |
→ |
3(1(5(1(0(0(x)))))) |
(15) |
|
3(0(1(0(x)))) |
→ |
3(1(5(2(0(0(x)))))) |
(16) |
|
3(0(1(0(x)))) |
→ |
3(1(5(5(0(0(x)))))) |
(17) |
|
3(0(1(0(x)))) |
→ |
3(2(2(1(0(0(x)))))) |
(18) |
|
3(0(1(0(x)))) |
→ |
3(5(1(0(0(2(x)))))) |
(19) |
|
3(0(1(0(x)))) |
→ |
3(5(1(5(0(0(x)))))) |
(20) |
|
3(0(1(0(x)))) |
→ |
5(1(1(3(0(0(x)))))) |
(21) |
|
3(4(1(0(x)))) |
→ |
3(1(2(4(0(x))))) |
(22) |
|
3(4(1(0(x)))) |
→ |
3(1(4(0(2(x))))) |
(23) |
|
3(4(1(0(x)))) |
→ |
3(1(5(4(0(x))))) |
(24) |
|
3(4(1(0(x)))) |
→ |
3(4(2(1(0(x))))) |
(25) |
|
3(4(1(0(x)))) |
→ |
3(1(1(5(4(0(x)))))) |
(26) |
|
3(4(1(0(x)))) |
→ |
3(1(2(1(4(0(x)))))) |
(27) |
|
3(4(1(0(x)))) |
→ |
3(1(2(5(4(0(x)))))) |
(28) |
|
3(4(1(0(x)))) |
→ |
3(1(4(2(0(2(x)))))) |
(29) |
|
3(4(1(0(x)))) |
→ |
3(1(5(4(0(2(x)))))) |
(30) |
|
3(4(1(0(x)))) |
→ |
3(1(5(5(4(0(x)))))) |
(31) |
|
3(4(1(0(x)))) |
→ |
3(4(2(1(1(0(x)))))) |
(32) |
|
3(4(1(0(x)))) |
→ |
3(4(5(1(2(0(x)))))) |
(33) |
|
0(1(4(1(0(x))))) |
→ |
0(1(1(4(0(2(x)))))) |
(34) |
|
0(2(0(1(0(x))))) |
→ |
0(2(0(0(3(1(x)))))) |
(35) |
|
0(2(0(1(0(x))))) |
→ |
2(0(0(0(3(1(x)))))) |
(36) |
|
0(3(0(1(0(x))))) |
→ |
0(0(3(1(3(0(x)))))) |
(37) |
|
0(3(0(1(0(x))))) |
→ |
0(0(3(3(1(0(x)))))) |
(38) |
|
0(3(0(1(0(x))))) |
→ |
0(0(3(5(1(0(x)))))) |
(39) |
|
0(3(0(1(0(x))))) |
→ |
2(0(0(3(1(0(x)))))) |
(40) |
|
0(3(4(1(0(x))))) |
→ |
0(2(0(4(3(1(x)))))) |
(41) |
|
0(5(0(1(0(x))))) |
→ |
0(0(0(1(5(2(x)))))) |
(42) |
|
0(5(0(1(0(x))))) |
→ |
0(0(1(5(1(0(x)))))) |
(43) |
|
0(5(0(1(0(x))))) |
→ |
0(2(0(0(1(5(x)))))) |
(44) |
|
3(0(1(0(0(x))))) |
→ |
3(1(3(0(0(0(x)))))) |
(45) |
|
3(0(1(1(0(x))))) |
→ |
3(1(0(1(2(0(x)))))) |
(46) |
|
3(0(2(1(0(x))))) |
→ |
2(0(3(1(1(0(x)))))) |
(47) |
|
3(0(2(1(0(x))))) |
→ |
2(3(1(5(0(0(x)))))) |
(48) |
|
3(0(2(1(0(x))))) |
→ |
3(1(2(0(1(0(x)))))) |
(49) |
|
3(0(2(1(0(x))))) |
→ |
3(1(2(0(5(0(x)))))) |
(50) |
|
3(0(5(1(0(x))))) |
→ |
3(1(5(2(0(0(x)))))) |
(51) |
|
3(1(0(1(0(x))))) |
→ |
2(0(3(1(1(0(x)))))) |
(52) |
|
3(1(0(1(0(x))))) |
→ |
3(1(1(1(0(0(x)))))) |
(53) |
|
3(1(0(1(0(x))))) |
→ |
3(1(2(1(0(0(x)))))) |
(54) |
|
3(1(4(1(0(x))))) |
→ |
3(1(2(1(4(0(x)))))) |
(55) |
|
3(1(4(1(0(x))))) |
→ |
3(1(5(1(4(0(x)))))) |
(56) |
|
3(2(0(1(0(x))))) |
→ |
0(2(3(1(5(0(x)))))) |
(57) |
|
3(2(0(1(0(x))))) |
→ |
2(0(3(1(1(0(x)))))) |
(58) |
|
3(3(0(1(0(x))))) |
→ |
3(1(2(0(3(0(x)))))) |
(59) |
|
3(3(0(1(0(x))))) |
→ |
3(1(2(3(0(0(x)))))) |
(60) |
|
3(3(4(1(0(x))))) |
→ |
3(1(2(4(3(0(x)))))) |
(61) |
|
3(3(4(1(0(x))))) |
→ |
3(1(3(4(0(2(x)))))) |
(62) |
|
3(3(4(1(0(x))))) |
→ |
3(1(4(3(1(0(x)))))) |
(63) |
|
3(4(0(1(0(x))))) |
→ |
0(2(4(1(3(0(x)))))) |
(64) |
|
3(4(0(1(0(x))))) |
→ |
3(1(4(0(0(2(x)))))) |
(65) |
|
3(4(0(1(0(x))))) |
→ |
3(2(0(4(1(0(x)))))) |
(66) |
|
3(4(4(1(0(x))))) |
→ |
3(1(1(4(4(0(x)))))) |
(67) |
The underlying signature is as follows:
{0/1, 1/1, 2/1, 3/1, 4/1, 5/1}Property / Task
Prove or disprove confluence.Answer / Result
No.Proof (by csi @ CoCo 2021)
1 Non-Joinable Fork
The system is not confluent due to the following forking derivations.
| t0
|
= |
0(0(1(0(x)))) |
|
→
|
0(2(0(0(3(1(x)))))) |
|
= |
t1
|
| t0
|
= |
0(0(1(0(x)))) |
|
→
|
0(2(0(4(1(0(x)))))) |
|
= |
t1
|
The two resulting terms cannot be joined for the following reason:
- When applying the cap-function on both terms (where variables may be treated like constants)
then the resulting terms do not unify.