A truth table shows that the result is always true (=1). Add another Boolean p so that the truth table shows all possible (p∧q)→(q→r)
| p |
q |
q̄ |
r |
r̄ |
p∧q̄ |
q→r̄ |
(p∧q̄)→(q→r̄) |
| T |
T |
F |
T |
F |
F |
F |
T |
| T |
T |
F |
F |
T |
F |
T |
T |
| T |
F |
T |
T |
F |
T |
T |
T |
| T |
F |
T |
F |
T |
T |
T |
T |
| F |
T |
F |
T |
F |
F |
F |
T |
| F |
T |
F |
F |
T |
F |
T |
T |
| F |
F |
T |
T |
F |
F |
T |
T |
| F |
F |
T |
F |
T |
F |
T |
T |
No matter whether p is TRUE or FALSE, the result is TRUE for every combination. So the given operation always results in TRUE. All inputs give TRUE as the result, so the given statement is always true.