## Content

### Deduction

We talked about the reasoning process of deduction earlier (Reasoning Retreat) and now is the time to look at its truth table (Table 8). This comes with perhaps a little surprise or two.

$p$ | $q$ | $p\to q$ |

0 | 0 | 1 |

0 | 1 | 1 |

1 | 0 | 0 |

1 | 1 | 1 |

It seems reasonable to think that if $p$ and $q$ were both true that you would get a correct implication proposition; hence the entry 1 in that part of the table. But some of the other entries may be a little mystifying, so let's go through them one by one.

So what if $p$ and $q$ are both false. How can we use a true argument to go from $p$ to $q$? Suppose that $p$ is 'Alyssa was prime minister of Australia in 2016' and $q$ is 'Alyssa was elected to parliament before 2016'. Now neither $p$ nor $q$ is true but $p \to q$ may be true, to be a prime minister of Australia you have to be elected first. Because there is the chance of the implication being true we take that as the preferred option.

So what if $p$ is false and $q$ is true. What can be said about a deduction from $p$ to $q$? An example using words is not easy to find but look at $p$ is '5 = 7'and $q$ is '0 = 0'. Then $p$ is false and $q$ is true, but if we multiply both sides of $p$ by 0 we correctly get $q$.

Finally, if $p$ is true and $q$ is false we would expect $p \to q$ to be false. If not we would have to rethink all of our known theorems!

Now in many ways this is unsatisfactory. It is possible that we might come up with counterexamples to some of the cases in the truth table. In that case it might be better to think of the truth here as founded on the fact that it is possible for a true implication to take us between $p$ and $q$ except in the one case where $p$ is true and $q$ is false.

Questions

- Make up your own examples to show that the truth table for deduction is correct.
- Let $p$, $q$ and $r$ be three propositions. Produce truth tables for the following:
- $\hspace{9px}p'\to q$;
- $\hspace{5px}p \to qr$;
- $p + q \to p'$;
- $\hspace{2px}(p + q') \to pq$;
- $\hspace{6px}(q \to p) + (p \to q)$.

- What is the truth table for $p$ if and only if $q$?
- Can you find a proposition that has the same truth table as that of $p \to q$?