(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
