Logic question in propositional calculus

Question

How do we prove the following formula for all natural numbers $n$ in propositional calculus

$[(q_{1}\vee q_{2}...q_{n})\wedge((q_{1}\Longrightarrow r)\wedge(q_{2}\Longrightarrow r)...(q_{n}\Longrightarrow r))]\Longrightarrow r$

Answer 1

Induction works here. If $n=1$, then you have $$ (q_1\wedge (q_1\rightarrow r))\rightarrow r. $$ This is an application of modus ponens. Then you can assume the truth of the statement for $n-1$ and prove the statement for $n$ by using or-elimination: You can split up $q_1\vee q_2\vee \cdots \vee q_n$ into $$ (q_1\vee\cdots \vee q_{n-1})\vee q_n. $$

EDIT: In answer to the question in the comments: You need both or-elimination and modus ponens to prove that. Use or-elimination and you have to prove two statements: $(q_1\wedge (q_1\to r))\to r$ and $(q_2\wedge (q_2\to r))\to r$. Each of these is proved using modus ponens.