# I Logic and distributive laws

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1. May 1, 2016

### The Subject

Im just reading this one example and i am stumped at this one step.

$$(R\to C) \land (S \to C) \\ (\neg R\lor C) \land (\neg S \lor C) \ \ \ \ \ \textrm{by conditional law}\\ (\neg R\land \neg S) \lor C \ \ \ \ \textrm{by distributive law}$$

I don't understand how it went from the second step to the third

my attempt from the second step was:
$$(\neg R \land \neg S) \lor (\neg R \land C) \lor (C \land \neg S) \lor C$$
but don't know where to go from here.

Did I do correctly applied the distributive law?

2. May 1, 2016

### zinq

The distributive laws in the propositional calculus say that

A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C)​

and also what you get when the symbols ∧ and ∨ are interchanged:

A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C).​

The A, B, C can of course stand for any propositions at all. Now you can probably see how choosing the appropriate A, B, C from your problem will get you from step 2 to step 3.

3. May 1, 2016

### Stephen Tashi

They could have said "by distributive law used in reverse", but actually there is no "forward" and "reverse" direction to the distributive law. It a human tendency to think what we go from the left hand side of an equivalence to the right right hand side, but you don't have to use equivalences that way. Do you see how to go from the third line back to the second line using the distributive law?

I think you tried to do more than apply the distributive law to the second step.