proving direct proofs without using truth tables Use a direct proof to prove the conditional statement:
$$\left[P\Rightarrow(Q\lor R)\right]\Rightarrow [(P\land \lnot R)\Rightarrow Q]$$
The symbol $\Rightarrow$ means "if/then", the symbol $\land$ means "and", the symbol $\lor$ means "or", and $\lnot$ means "not"
I can understand how to do the steps from one conditional statement to the other, my problem is the reasoning behind each step.
 A: Suppose $P \Rightarrow (Q \vee R)$ is a true statement.  Suppose $(P \wedge \sim R)$ is a true statement. (I apologize for using a different notation for "not.")  We want to show that $Q$ is true.  
Since $P$ is true, either $R$ is true or $Q$ is true. But we assumed $R$ is not true, so we conclude that $Q$ is true.
Thus the statement $P\wedge \sim R \Rightarrow Q$ is true.
A: $$[P \rightarrow  (Q \vee R)] \rightarrow [(P \wedge \neg R) \rightarrow Q]$$
Let's prove this with a contradiction. If you don't know, a proof by contradiction works like this:


*

*Assume statement -A

*Show that this assumption leads to an impossibility.

*Conclude that since -A leads to an impossibility, then it can't be true. Therefore, A must be true


Our assumption is going to be:
Assume 
$$[P \rightarrow  (Q \vee R)] \rightarrow [(P \wedge \neg R) \rightarrow Q]$$
is a false statement. Therefore, there must exist values such that
$[P \rightarrow  (Q \vee R)]$ is a 1 (i.e. true) and
$[(P \wedge \neg R) \rightarrow Q]$ is a 0 (i.e. false)
In order to make $[(P \wedge \neg R) \rightarrow Q]$ be false:


*

*P must be 1

*-R must be 1 (i.e. R must be 0)

*Q must be 0


Now we know from our assumption that $[P \rightarrow  (Q \vee R)]$ is a 1. However, if we input the values for P, Q, and R that we got above, then
$[P \rightarrow  (Q \vee R)]$ evaluates to 0 (it should've evaluated to 1 because of our assumption). This is an impossibility. Therefore, we have reached a contradiction. That means that our initial assumption that 
$$[P \rightarrow  (Q \vee R)] \rightarrow [(P \wedge \neg R) \rightarrow Q]$$
is false is incorrect. Therefore, the statement is true.
