Let's make a truth table containing all four statements.
| \(P\) | \(Q\) | \(R\) | \(P \imp R\) | \(Q \imp R\) | \(P \vee Q\) |
| T | T | T | T | T | T |
| T | T | F | F | F | T |
| T | F | T | T | T | T |
| T | F | F | F | T | T |
| F | T | T | T | T | T |
| F | T | F | T | F | T |
| F | F | T | T | T | F |
| F | F | F | T | T | F |
Look at the second to last row. Here all three premises of the argument are true, but the conclusion is false. Thus this is not a valid deduction rule.
While we have the truth table in front of us, look at rows 1, 3, and 5. These are the only rows in which all of the statements \(P \imp R\text{,}\) \(Q \imp R\text{,}\) and \(P\vee Q\) are true. It also happens that \(R\) is true in these rows as well. Thus we have discovered a new deduction rule we know is valid:
| \(P \imp R\) | |
| \(Q \imp R\) | |
| \(P \vee Q\) | |
| \(\therefore\) | \(R\) |