True or false, both Weak and Strong Leibniz can be used in a proof for $p \lor q \iff (\forall y)(y = fx \land (\neg p \land q)) = (\forall y)(y = fx \land (\neg(p \lor \neg q)))$ since conditional and unconditional substitution will be equivalent in this case?
Added by Michael J.
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Mark each sentence as true or false, where $p, q,$ and $r$ are arbitrary statements, $t$ a tautology, and $f$ a contradiction. $$ \sim(p \vee q)=\sim p \vee \sim q $$
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Mark each sentence as true or false, where $p, q,$ and $r$ are arbitrary statements, $t$ a tautology, and $f$ a contradiction. $$ \sim(p \wedge q) \equiv \sim p \wedge \sim q $$
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