We consider the TRS containing the following rules:
0(1(2(3(4(x))))) | → | 0(2(1(3(4(x))))) | (1) |
0(5(1(2(4(3(x)))))) | → | 0(5(2(1(4(3(x)))))) | (2) |
0(5(2(4(1(3(x)))))) | → | 0(1(5(2(4(3(x)))))) | (3) |
0(5(3(1(2(4(x)))))) | → | 0(1(5(3(2(4(x)))))) | (4) |
0(5(4(1(3(2(x)))))) | → | 0(5(4(3(1(2(x)))))) | (5) |
The underlying signature is as follows:
{0/1, 1/1, 2/1, 3/1, 4/1, 5/1}