(p ˄ q) <-> ~q
Start with this
p|q|
T|T|
T|F|
F|T|
F|F|
The simplest operation that "(p ˄ q) <-> ~q" contains is the "~q",
So make that the heading of the next column:
p|q|~q|
T|T| |
T|F| |
F|T| |
F|F| |
Now fill in the column by looking at the column for q,
and remer the rule: If q has a T, ~q will have an F, and
vice-versa:
p|q|~q|
T|T| F|
T|F| T|
F|T| F|
F|F| T|
The next simplest operation that (p ˄ q) <-> ~q" contains is the "(p ˄ q)",
So make that the heading of the next column:
p|q|~q|(p ˄ q)|
T|T| F| |
T|F| T| |
F|T| F| |
F|F| T| |
Now fill in that column by looking at the columns for p and for q,
and remembering the rule for p ˄ q: "T ˄ T" gets a T, but anything
else gets a F:
p|q|~q|(p ˄ q)|
T|T| F| T |
T|F| T| F |
F|T| F| F |
F|F| T| F |
The next simplest operation that "(p ˄ q) <-> ~q" contains is the only
thing left, the WHOLE THING, "(p ˄ q) <-> ~q"
p|q|~q|(p ˄ q)|(p ˄ q) <-> ~q
T|T| F| T |
T|F| T| F |
F|T| F| F |
F|F| T| F |
Now we fill in that final column by looking at the columns for "(p ˄ q)|"
and for ~q, and remembering the rule for <=>: T<->T and F<->F gets
T and the others get F
p|q|~q|(p ˄ q)|(p ˄ q) <-> ~q
T|T| F| T | F
T|F| T| F | F
F|T| F| F | T
F|F| T| F | F
Edwin
