Will a contradiction always be false?

Will a contradiction always have only false values?

A tautology will always have true vales, is the opposite true for contradictions?

For example: $(p \ \& \ q) \ \& \ (p \ \& \ \neg q)$ has the truth table:

• YES : a contradiction is a formula that is always false, like $p \land \lnot p$ – Mauro ALLEGRANZA Oct 2 '16 at 18:30
• Yes: That it is always false is in fact the definition of a contradiction. – lemontree Oct 2 '16 at 19:20
• @lemontree yup, and if you look at the table of 16 operators, you'll see F aka contradiction with false in every row. – The Great Duck Oct 2 '16 at 22:56
• See the principal of explosion en.m.wikipedia.org/wiki/Principle_of_explosion – Awn Jun 4 '18 at 18:46

$a \land b\land c \land .... \land p\land \lnot p$ will never be true, because $p \land \lnot p$ is False, and no matter what you "and" to that to form a new statement, it will still be false. Any statement for which no matter the assignment of truth-values to the propositions, ends in the evaluation of False, is a contradiction.
• Note also that a tautology is always true. For example, say we have $p\lor q\lor r \lor \lnot p$. Here, the truth-value of q and that of r are irrelevant to the truth value of the proposition. $p\lor \lnot p$ is always true, because $p$ is true, or $\lnot p$ is true. If any proposition is transformed to a truth-table has ALL T's in the last column, you can be assured the proposition is a tautology.: – Namaste Oct 2 '16 at 19:33
• What if the "..." in a $\land$ b $\land$ c $\land$ ... p $\land$ $\lnot$p consisted of a sequence with more variables and $\land$ symbols than could ever get written in the physical universe? In such a case, the assignment of truth values to the propositions would never end. Though, evaluating from right to left we would quickly reach a falsity, making the entire conjunction false. – Doug Spoonwood Oct 2 '16 at 21:57