(from Boolean algebra: 0 + 0 = 0, 0 + 1 = 1 + 0 = 1, 1 + 1 = 1) & (logical AND) between propositions; in the minterms AND is omitted in a manner similar to arithmetic multiplication: e.g. x'y'z =defined ~x & ~y & z (From Boolean algebra: 0·0 = 0, 0·1 = 1·0 = 0, 1·1 = 1, where "·" is shown for clarity) → (logical IMPLICATION): read as IF ... THEN ..., or " IMPLIES ", P → Q = defined NOT P OR Q Given a proposed conclusion such as "No X is a Z", one can test whether or not it is a correct deduction by use of a truth table. The easiest method is put the starting formula on the left (abbreviate it as P) and put the (possible) deduction on the right (abbreviate it as Q) and connect the two with logical implication i.e. P → Q, read as IF P THEN Q. If the evaluation of the truth table produces all 1s under the implication-sign (→, the so-called major connective) then P → Q is a tautology. Given this fact, one can "detach" the formula on the right (abbreviated as Q) in the manner described below the truth table. Given the example above, the formula for the Euler and Venn diagrams is: "No Ys are Zs" and "All Xs are Ys": ( ~(y & z) & (x → y) ) =defined P And the proposed deduction is: "No Xs are Zs": ( ~ (x & z) ) =defined Q So now the formula to be evaluated can be abbreviated to: ( ~(y & z) & (x → y) ) → ( ~ (x & z) ): P → Q IF ( "No Ys are Zs" and "All Xs are Ys" ) THEN ( "No Xs are Zs" ) At this point the above implication P → Q (i.e. ~(y & z) & (x → y) ) → ~(x & z) ) is still a formula, and the deduction – the "detachment" of Q out of P → Q – has not occurred. But given the demonstration that P → Q is tautology, the stage is now set for the use of the procedure of modus ponens to "detach" Q: "No Xs are Zs" and dispense with the terms on the left. Modus ponens (or "the fundamental rule of inference") is