Discrete Maths Logic Question p = False, q = True and r = False.
Is $¬(p∨q)∧(¬p∨r)$ = false?
My reasoning:
$$(p∨q)=T \text{ as it is  (F or T)}$$
but its the negation so $¬(p∨q)=F$?  
Then, $(¬p∨r)$ as p is F but its the negation again so its T and r=F.
So its $(T\lor F)$ in this case only one has to be true because $\lor$ means or,
so it is T. Now we have $F \land T$ which comes to False because $\land$='and' and both needs to be true to make it true. That's how I came to false. 
Is this correct? My lecturer still has not put up the answer sheet and never responds to E-mails....
 A: Note that simply having successfully worked out the fact that $\lnot(p \lor q)$ is false, you are effectively done. 
Why? Because in any conjunction, say $A\land B$, (here, $A, B$ may be any proposition),
by the truth-table definition of conjunction, $A \land B$ is true if and only if both $A$ is true and $B$ is true.
If $A$ is false, then $A\land B$ is necessarily false. Likewise, if we know only that $B$ is false, then we know $A\land B$ is necessarily false.
So $$\underbrace{\underbrace{\lnot (\underbrace{p\lor q}_{\text{T}})}_{\text{F}} \land (\lnot p \lor q)}_{\text {F}}$$
A: This question can be answered in two ways. 
1) Using Binary Mathematics, Too simple, use 
1 for true,
0 for false,
+ for ∨, 
. for ∧, 

then 
¬(p∨q)∧(¬p∨r) given that p = false, q = true, r = false
   can be rewritten as:
¬(0+1).(¬0+0) = ¬1.(1+0)
¬1.(1+0) = 0.1 
0.1 = 0

as we took 0 for false. the end answer is false.
2) The real proof: using Mathematical Logic
¬(p∨q)∧(¬p∨r)

= ¬(F∨T)∧(¬F∨F)      { substituting     p=false(F), q=true(T), r=false(F) }
= ¬(T)∧(¬F∨F)        { as F V T = F}
= F∧(T∨F)            { as ¬T=F and ¬F=T}
= F                  { as F∧(anything on this side) will be F}

Hence the answer is False.
