Construct truth tables and test validity

Truth values belong to propositions

Section 1D established that a proposition has a definite truth value. Here P, Q and R stand for arbitrary propositions. Use 1 for true and 0 for false. A truth table lists every possible assignment to the input propositions; it does not assume that every row describes the same real situation

Not, and, and inclusive or

Negation reverses a value. A conjunction is true only when both inputs are true. Inclusive or is true when at least one input is true, including both. Evaluate brackets from the inside out

P and Q are both true. What is the truth value of their inclusive disjunction?

Exclusive or differs in one row

Exclusive or means exactly one input is true. The symbol used here is ⊕. Its values for input rows (1,1), (1,0), (0,1), (0,0) are 0, 1, 1, 0. It differs from inclusive or only when both inputs are true. Ordinary language can be ambiguous, so state which meaning is intended

P and Q are both true. What is the truth value of P exclusive-or Q?

Conditional and biconditional

P implies Q fails only when P is true and Q is false. With P false, the conditional is true: that row supplies no counterexample. P if and only if Q is true when the two truth values agree. A true conditional does not assert that its hypothesis is true

P is false and Q is false. What is the truth value of P implies Q?

Build every row exactly once

For two independent input propositions list 11, 10, 01 and 00. For three, list 111, 110, 101, 100, 011, 010, 001 and 000. There are two to the power of n assignments for n inputs because each can independently be true or false. Add intermediate columns for nested expressions before evaluating the final column

Worked compound evaluation

Let P be true and Q false. To evaluate not(P and Q) if and only if (not P or not Q), compute the two sides separately. P and Q is false, so the left side is true. Not P is false and not Q is true, so the right side is true. The biconditional of true and true is true. One successful row is not yet an equivalence proof

A formula has final-column values 1, 0, 1, 0 across all four assignments. How is it classified?

Equivalence is agreement in every row

Two formulas are logically equivalent when their final columns agree for every assignment. Equivalently, their biconditional is a tautology. A contradiction is false in every row. Complete truth tables prove propositional identities because they exhaust every assignment; checking a few numerical examples of a quantified statement does not exhaust an infinite domain

Construct an equivalence proof, ungraded

Construct all four rows for not(P or Q) and (not P and not Q). Include separate columns for the inner expressions. Decide whether the two formulas are equivalent and justify your answer using every row. Independent construction and explanation are ungraded

Model: negating an inclusive disjunction

In row order 11, 10, 01, 00, P or Q is 1, 1, 1, 0, so not(P or Q) is 0, 0, 0, 1. The pairs (not P, not Q) are 00, 01, 10, 11; their conjunction is also 0, 0, 0, 1. All four assignments agree, proving equivalence. Replacing the conjunction by a disjunction would fail, for example at P = 1 and Q = 0.

Validity tests the connection to the conclusion

An argument is valid if no assignment makes all premises true and the conclusion false. Combine its premises using and, then form a conditional to the conclusion. The argument is valid exactly when that conditional is a tautology. Rows with a false premise do not refute validity. Soundness additionally requires that the premises are actually true in the intended interpretation

For an argument with premises P implies Q and Q, proposed conclusion P, which assignment refutes validity?

Valid patterns and tempting reversals

Modus ponens uses P implies Q and P to conclude Q. Modus tollens uses P implies Q and not Q to conclude not P. Affirming Q does not establish P, and denying P does not establish not Q. Inclusive or with not P can establish Q; inclusive or with P cannot establish not Q.

Test and repair an argument, ungraded

A claim uses premises “if an integer is divisible by 10, then it is divisible by 5” and “this integer is divisible by 5” to conclude that it is divisible by 10. Refute the argument with a permitted integer, identify its propositional pattern, and give a valid modus-tollens argument using the same conditional. Written construction is ungraded

Model: a counterexample and a valid repair

The integer 15 is divisible by 5 but not by 10. It makes P false and Q true, so both premises hold while the conclusion fails: affirming the consequent is invalid. A valid repair is: if an integer is divisible by 10, it is divisible by 5; this integer is not divisible by 5; therefore it is not divisible by 10. This is modus tollens. The original conditional is true because ten times any integer is five times twice that integer

An argument has inconsistent premises: P and not P. What can a truth table establish about a proposed conclusion Q?