(P & Q) <=> ~(P -> ~Q)

Using the rules:

Under P put TTFF,

Under Q put TFTF,

The rule for "~" is "~T is F and ~F is T",

The rule for "&" is "only T&T is T, all others F",

The rule for "V" is "only FVF is F, all others T",

The rule for "->" is "only T->F is F, all other T",

The rule for "<->" is "only T<->T and F<->F are T, all others F,

make this truth table:

| P | Q | ~Q | P & Q | P -> ~Q | ~(P -> ~Q) | (P&Q) <-> ~(P -> ~Q} |
| T | T |  F |   T   |    F    |     T      |        T             | 
| T | F |  T |   F   |    T    |     F      |        T             |
| F | T |  F |   F   |    T    |     F      |        T             |
| F | F |  T |   F   |    T    |     F      |        T             |

The proposition is proved because there are only T's in the last 
column.

Therefore we can replace the biconditional symbol <->, by the
stronger equivalence symbol <=> and write

(P&Q) <=> ~(P -> ~Q}

Edwin