Determine whether the compound propositions in 63–64 are satisfiable. 63. (¬p ∨ ¬q) ∧ (p -> q)
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A compound proposition is said to be satisfiable if we can assign the variables (propositions), within the compound proposition, truth values that allow the entire formula to evaluate to TRUE. For example, the proposition P ∧ Q is said to be satisfiable because if we set P to TRUE and Q to TRUE, then the resulting value is true. Determine whether the following compound propositions are satisfiable, and if so, define the truth values. Otherwise, prove that it is not. a. (P ↔ Q) ∧ ( P ↔ Q) b. (P ∨ P) ⊕ (( P ∨ R ∨ S) ∧ ( P ∨ T ∨ S) ∧ (P ∨ Q ∨ R)) c. ( P ∨ Q ∨ R) ∧ ( P ∨ Q ∨ S) → Q
Adi S.
Determine whether each of these compound propositions is satisfiable. a) $(p \vee \neg q) \wedge(\neg p \vee q) \wedge(\neg p \vee \neg q)$ b) $(p \rightarrow q) \wedge(p \rightarrow \neg q) \wedge(\neg p \rightarrow q) \wedge(\neg p \rightarrow \neg q)$ c) $(p \leftrightarrow q) \wedge(\neg p \leftrightarrow q)$
The Foundations: Logic and Proofs
Propositional Equivalences
Determine whether each of these compound propositions is satisfiable. $$ \begin{array}{l}{\text { a) }(p \vee q \vee \neg r) \wedge(p \vee \neg q \vee \neg s) \wedge(p \vee \neg r \vee \neg s) \wedge} \\ {(\neg p \vee \neg q \vee \neg s) \wedge(p \vee q \vee \neg s)} \\ {\text { b) }(\neg p \vee \neg q \vee r) \wedge(\neg p \vee q \vee \neg s) \wedge(p \vee \neg q \vee \neg s) \wedge} \\ {(\neg p \vee \neg r \vee \neg s) \wedge(p \vee q \vee \neg r) \wedge(p \vee \neg r \vee \neg s)} \\ {(\neg p \vee q \vee r \vee \neg s) \wedge(p \vee q \vee \neg r) \wedge(q \vee \neg r \vee s) \wedge(\neg p \vee} \\ {r \vee s ) \wedge(\neg p \vee q \vee \neg s) \wedge(p \vee \neg q \vee \neg r) \wedge(\neg p \vee} \\ {\neg q \vee s ) \wedge(\neg p \vee \neg r \vee \neg s)}\end{array} $$
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