We consider the TRS containing the following rules:
| and3(x,y,F) | → | F | (1) |
| and3(T,T,T) | → | T | (2) |
| and3(x,y,z) | → | and3(y,z,x) | (3) |
The underlying signature is as follows:
{and3/3, F/0, T/0}To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| and3(x,y,z) | → | and3(y,z,x) | (3) |
| and3(T,T,T) | → | T | (2) |
| and3(x,y,F) | → | F | (1) |
| and3(y,F,x) | → | F | (4) |
All redundant rules that were added or removed can be simulated in 3 steps .
To prove that the TRS is (non-)confluent, we show (non-)confluence of the following modified system:
| and3(y,F,x) | → | F | (4) |
| and3(x,y,F) | → | F | (1) |
| and3(T,T,T) | → | T | (2) |
| and3(x,y,z) | → | and3(y,z,x) | (3) |
| and3(F,x,y) | → | F | (5) |
| and3(F,z,x) | → | F | (6) |
All redundant rules that were added or removed can be simulated in 2 steps .
Confluence is proven using the following terminating critical-pair-closing-system R:
| and3(F,x,y) | → | F | (5) |
| and3(y,F,x) | → | F | (4) |
| and3(T,T,T) | → | T | (2) |
| and3(x,y,F) | → | F | (1) |
| [and3(x1, x2, x3)] | = | 2 · x1 + 6 · x2 + 1 · x3 + 0 |
| [F] | = | 5 |
| [T] | = | 3 |
| and3(F,x,y) | → | F | (5) |
| and3(y,F,x) | → | F | (4) |
| and3(T,T,T) | → | T | (2) |
| [and3(x1, x2, x3)] | = | 2 · x1 + 4 · x2 + 1 · x3 + 1 |
| [F] | = | 0 |
| and3(x,y,F) | → | F | (1) |
There are no rules in the TRS. Hence, it is terminating.