MathematicsMedium115×since 2008Q4374If the truth value of the statement (P∧(∼R))→((∼R)∧Q)(P \wedge(\sim R)) \rightarrow((\sim R) \wedge Q)(P∧(∼R))→((∼R)∧Q) is F, then the truth value of which of the following is F\mathrm{F}F ?AP∨Q→ ∼R\mathrm{P} \vee \mathrm{Q} \rightarrow \,\sim \mathrm{R}P∨Q→∼RBR∨Q→ ∼P\mathrm{R} \vee \mathrm{Q} \rightarrow \,\sim \mathrm{P}R∨Q→∼PC∼(P∨Q)→∼R\sim(\mathrm{P} \vee \mathrm{Q}) \rightarrow \sim \mathrm{R}∼(P∨Q)→∼RD∼(R∨Q)→ ∼P\sim(\mathrm{R} \vee \mathrm{Q}) \rightarrow \,\sim \mathrm{P}∼(R∨Q)→∼PCheck answerSkip