Contrast syntactic proof, which uses logical operations such as modus ponens to rearrange and combine theorems derived from the premisses, but which is not guaranteed to find an elegant solution.

Suppose we know that (A) I won't need my umbrella just in case it doesn't rain today, (B) it always rains on Saturdays, and (C) today is Saturday. We want to prove whether I'll need my umbrella. Let these propositions be built out of simple ones:

P = I'll need my umbrellaThen the propositions we know as premisses (know to be true; are assuming to be true) are:

Q = it will rain today

R = today is Saturday

(A) ~Q <-> ~PAnd what we want to ascertain is whether P (I'll need my umbrella) is a valid deduction from these.

(B) R -> Q

(C) R

Now we build the truth table for these:

P Q R ~Q <-> ~P , R -> Q T T T T T F T F T T F F F T T F T F F F T F F FBoolean operators inside complex propositions need to evaluated in order of bracketing or binding, innermost to outermost. So do the negatives first:

P Q R ~Q <-> ~P , R -> Q T T T F F T T F F F T F T T F T F F T F F T T F T F T F F T F F T T T F F F T TNow do the biconditional and the implication:

P Q R ~Q <-> ~P , R -> Q T T T F T F T T T F F T F T T F T T F F F T F F T F F T F T T F F T T F T F F F T T F F T T T T F F F F T T T TWe want to work out what follows if all three of our premisses are true. The first premiss (A), which is ~Q <-> ~P, is true in the following lines of the table:

P Q R ~Q <-> ~P , R -> Q T T T F T F T T T F F T F T F F T T T T F F F F T T T TThe next one, (B) which is R -> Q, is true in the following lines of that subtable:

P Q R ~Q <-> ~P , R -> Q T T T F T F T T T F F T F T F F F T T T TThe final premiss (C) is R, and that's true in the following part of the previous subtable:

P Q R ~Q <-> ~P , R -> Q T T T F T F TThe conclusion we're looking for is P. We see that in all lines of the current table, P is true. So P follows deductively from the three premisses.