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In page 37 of A Transition to Advanced Mathematics, the authors intend to prove a statement of general logical form $P\vee Q\Rightarrow R_{1}\vee R_{2}$ where $P,Q,R_{1},R_{2}$ are predicates, "...The proof method we choose is to show that $P\Rightarrow R_{1}$ and $Q\Rightarrow R_{2}$." I cannot understand how these are equivalent, particularly because of the Constructive Dilemma stated in page 16 of Set Theory With Applications: $$ [(P\Rightarrow Q)\wedge(R\Rightarrow S)]\Rightarrow(P\vee R\Rightarrow Q\vee S) $$ As you can observe, this is a conditional statement and not a biconditional which I interpret as one is not permitted to derive $(P\Rightarrow Q)\wedge (R\Rightarrow S)$ while assumming $(P\vee R\Rightarrow Q\vee S)$ as antecedent at all times. Can anyone explain how their "proof method" is valid?

mali1234
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    They are not *equivalent, but if they prove that $P \implies R_1$ and $Q \implies R_2$, then that implies* that $P \vee Q \implies R_1 \vee R_2$, which is what they want to show, right? – Joe Nov 21 '21 at 13:53
  • In the statement on p.$37$ of the cited book that $\ [P\vee Q\Rightarrow R]\iff[(P\Rightarrow R)\wedge(Q\Rightarrow R)]\ $ is a tautology, the logical expression should instead be $\ [P\vee Q\Rightarrow R]\Leftarrow[(P\Rightarrow R)\wedge(Q\Rightarrow R)]\ $. I suspect it's just a misprint. – lonza leggiera Nov 21 '21 at 14:09
  • The converse of what you say is true Joe. But as you have stated, they are not equivalent and you cannot assume the consequent and derive the antecedent. – mali1234 Nov 22 '21 at 13:19

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