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Korvikt [17]
4 years ago
8

Prove that P (P) = (QA ~ Q)] is a tautology.

Mathematics
1 answer:
alekssr [168]4 years ago
8 0

Answer:

The statement P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)] is a tautology.

Step-by-step explanation:

A tautology is a formula which is "always true" that is, it is true for every assignment of truth values to its simple components.

To show that this statement is a tautology we are going to use a table of logical equivalences:

P \leftrightarrow [(\lnot P) \rightarrow (Q \land \lnot Q)] \equiv

\equiv (P \land [(\lnot P)\rightarrow (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [(\lnot P)\rightarrow (Q \land \lnot Q)]) by the logical equivalences involving bi-conditional statements

\equiv (P \land [\lnot(\lnot P)\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [\lnot(\lnot P)\lor (Q \land \lnot Q)]) by the logical equivalences involving conditional statements

\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot [ P\lor (Q \land \lnot Q)]) by the Double negation law

\equiv (P \land [P\lor (Q \land \lnot Q)]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q)) by De Morgan's law

\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot(Q \land \lnot Q)) by the Negation law

\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor \lnot(\lnot Q)) by De Morgan's law

\equiv (P \land [P\lor F]) \lor(\lnot P \land \lnot P\land \lnot Q \lor  Q) by the Double negation law

\equiv (P \land P) \lor(\lnot P \land \lnot P\land \lnot Q \lor  Q) by the Identity law

\equiv (P) \lor(\lnot P \land \lnot P\land \lnot Q \lor  Q) by the Idempotent law

\equiv (P) \lor(\lnot P \land \lnot P\land  (Q\lor \lnot Q)) by the Commutative law

\equiv (P) \lor(\lnot P \land \lnot P\land T) by the Negation law

\equiv (P) \lor(\lnot (P \lor P)\land T) by De Morgan's law

\equiv (P) \lor(\lnot (P)\land T) by the Idempotent law

\equiv (P \lor\lnot P) \land(P \lor T) by the Distributive law

\equiv (T) \land(P \lor T) by the Negation law

\equiv (T) \land(T) by the Domination law

\equiv T

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Step-by-step explanation:

Let x_1 be the number of widgets sold and x_2 be the number of gizmos sold.

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