# Use the laws of logic to show that $[a\Rightarrow(b\lor c)]\Leftrightarrow[(a\land\lnot b)\Rightarrow c]$

I am trying to prove that $$[a\Rightarrow(b\lor c)]\Leftrightarrow[(a\land\lnot b)\Rightarrow c]$$.

My proof is the following:

• $$a\Rightarrow(b\lor c)~$$ Premise
• $$(a\Rightarrow b)\lor c~$$ Associative Law
• $$(\lnot a\lor b)\lor c~$$ Material Implication
• $$\lnot(a\lor\lnot b)\lor c~$$ De Morgan's Law
• $$(a\land\lnot b)\Rightarrow c~$$ Material Implication

I'm having doubts about my second step. I tried to check for the validity of my step using truth table, and the statements in the first and second steps are logically equivalent. Is my application of associative law legal?

• But you can use Material Implication directly on 1st line to get 3rd – Mauro ALLEGRANZA Jun 17 '20 at 12:29
• I'm slightly confused: commonly, associativity is something only concerning one type of operation not two; could you state the laws/interfer rules you used as their formulation might differ from source to source. – mrtaurho Jun 17 '20 at 12:34
• The associative law is a tad confusing. What is meant is $$a\Rightarrow (b\lor c) \equiv \neg a\lor (b\lor c) \equiv (\neg a \lor b) \lor c$$ – Alvin Lepik Jun 17 '20 at 13:02

$$a \rightarrow (b \lor c) \iff \neg a \lor (b \lor c) \iff (\neg a \lor b) \lor c$$