The rewrite relation of the following TRS is considered.
|
5(4(1(x1))) |
→ |
2(5(5(1(1(1(0(2(3(4(x1)))))))))) |
(1) |
|
4(1(1(3(x1)))) |
→ |
4(4(5(1(4(0(2(0(3(3(x1)))))))))) |
(2) |
|
4(4(1(3(x1)))) |
→ |
5(1(1(4(5(5(3(3(2(3(x1)))))))))) |
(3) |
|
0(0(4(1(4(x1))))) |
→ |
3(3(1(1(0(3(1(4(2(4(x1)))))))))) |
(4) |
|
0(5(0(4(2(x1))))) |
→ |
0(0(3(5(3(3(2(3(4(5(x1)))))))))) |
(5) |
|
2(4(3(4(0(x1))))) |
→ |
2(5(2(2(5(5(0(3(2(3(x1)))))))))) |
(6) |
|
4(4(2(3(4(x1))))) |
→ |
2(2(4(5(3(3(1(1(1(4(x1)))))))))) |
(7) |
|
5(2(4(3(0(x1))))) |
→ |
2(5(5(2(3(2(5(4(0(0(x1)))))))))) |
(8) |
|
0(0(0(2(0(1(x1)))))) |
→ |
3(3(5(3(1(3(1(4(0(1(x1)))))))))) |
(9) |
|
1(2(0(1(3(2(x1)))))) |
→ |
2(0(3(2(1(2(3(1(0(3(x1)))))))))) |
(10) |
|
1(3(5(4(3(5(x1)))))) |
→ |
1(3(3(1(1(5(1(0(1(5(x1)))))))))) |
(11) |
|
1(4(1(0(1(2(x1)))))) |
→ |
1(0(3(1(5(3(3(4(5(2(x1)))))))))) |
(12) |
|
3(2(4(3(5(5(x1)))))) |
→ |
3(5(5(5(1(4(4(5(4(5(x1)))))))))) |
(13) |
|
3(4(3(2(0(5(x1)))))) |
→ |
2(3(3(5(3(3(1(3(3(5(x1)))))))))) |
(14) |
|
3(5(5(5(0(2(x1)))))) |
→ |
3(3(5(1(0(0(3(3(1(0(x1)))))))))) |
(15) |
|
5(0(4(3(0(1(x1)))))) |
→ |
0(3(1(2(5(1(1(1(2(1(x1)))))))))) |
(16) |
|
5(5(4(3(2(4(x1)))))) |
→ |
5(4(2(2(5(2(1(2(3(4(x1)))))))))) |
(17) |
|
0(0(2(0(5(2(1(x1))))))) |
→ |
0(4(3(5(2(3(5(5(2(1(x1)))))))))) |
(18) |
|
0(1(3(4(2(1(2(x1))))))) |
→ |
0(1(2(0(3(1(2(5(2(2(x1)))))))))) |
(19) |
|
0(1(4(1(4(4(3(x1))))))) |
→ |
0(1(0(1(1(2(0(5(2(3(x1)))))))))) |
(20) |
|
0(5(1(1(3(4(1(x1))))))) |
→ |
0(2(2(0(2(3(1(5(5(3(x1)))))))))) |
(21) |
|
1(3(1(0(4(1(0(x1))))))) |
→ |
1(3(3(3(3(0(0(3(1(0(x1)))))))))) |
(22) |
|
1(3(2(0(1(2(1(x1))))))) |
→ |
5(2(4(2(4(4(2(5(1(3(x1)))))))))) |
(23) |
|
1(3(5(4(4(1(2(x1))))))) |
→ |
2(0(5(0(2(3(2(2(2(2(x1)))))))))) |
(24) |
|
1(4(0(0(2(5(0(x1))))))) |
→ |
1(0(3(2(2(2(1(0(5(0(x1)))))))))) |
(25) |
|
2(0(1(5(1(4(2(x1))))))) |
→ |
2(2(5(5(2(1(2(4(0(5(x1)))))))))) |
(26) |
|
2(0(4(0(4(3(0(x1))))))) |
→ |
3(1(3(3(4(5(2(3(0(3(x1)))))))))) |
(27) |
|
3(0(4(1(0(0(0(x1))))))) |
→ |
5(2(3(5(4(5(5(1(3(0(x1)))))))))) |
(28) |
|
3(0(4(3(0(3(4(x1))))))) |
→ |
3(5(3(1(0(3(2(3(0(4(x1)))))))))) |
(29) |
|
3(4(0(4(4(2(3(x1))))))) |
→ |
3(2(4(4(5(1(1(4(5(0(x1)))))))))) |
(30) |
|
3(4(4(1(4(1(1(x1))))))) |
→ |
4(2(0(3(2(1(3(0(0(1(x1)))))))))) |
(31) |
|
3(5(0(4(1(5(0(x1))))))) |
→ |
2(3(1(0(5(4(3(0(3(3(x1)))))))))) |
(32) |
|
3(5(0(5(5(5(3(x1))))))) |
→ |
3(3(5(4(0(1(4(5(3(0(x1)))))))))) |
(33) |
|
4(1(5(2(0(1(5(x1))))))) |
→ |
2(5(4(2(3(3(0(0(0(5(x1)))))))))) |
(34) |
|
4(3(0(1(2(4(2(x1))))))) |
→ |
4(3(3(1(1(0(3(2(2(5(x1)))))))))) |
(35) |
|
4(3(4(1(4(4(3(x1))))))) |
→ |
1(4(3(3(3(1(3(1(2(2(x1)))))))))) |
(36) |
|
5(0(4(2(0(4(2(x1))))))) |
→ |
5(2(2(0(1(5(5(2(4(2(x1)))))))))) |
(37) |
|
5(0(4(2(5(5(0(x1))))))) |
→ |
0(3(3(1(2(3(0(0(2(3(x1)))))))))) |
(38) |
|
5(4(1(2(0(4(1(x1))))))) |
→ |
4(5(4(2(2(3(4(5(5(3(x1)))))))))) |
(39) |
There are 234 ruless (increase limit for explicit display).
As carrier we take the set
{0,...,5}.
Symbols are labeled by the interpretation of their arguments using the interpretations
(modulo 6):
There are 1404 ruless (increase limit for explicit display).
There are 1404 ruless (increase limit for explicit display).
There are no rules in the TRS. Hence, it is terminating.