Question

Help with symbolic logic:
Simplify:
(~r∧p)∧(s∨r)∧(~t∨p)∧(t∨~s)

Where p, r,s and t are propositions
∧ means conjunction
∨ means disjunction
~ means negation.

Answer 1

Are you familiar with truth tables? @Juarismi

Answer 2

Yes, and i hate them.

Answer 3

~R & P
& (S | R) -> so far we're true if ~R & P & S
& (~T | P) -> so far we're true if ~R & P & S
& (T | ~S) -> so far we're true if ~R & P & S & T

Answer 4

just go through each step and logically consider what is redundant... first step we know ~R & (S | R) reduces to ~R & S, second step we know P & (~T | P) reduces to P. for S & (T | ~S) we know that this just reduces to T.

Answer 5

man I miss studying propositional logic. We had to do a bunch of proofs using our tautologies and inference rules in MEGSSS.

Answer 6

looks it ends with ~R & P & S

Answer 7

@ganeshie8
http://www.wolframalpha.com/input/?i=%28~r%26p%29%26%28s%7Cr%29%26%28~t%7Cp%29%26%28t%7C~s%29
here you can see the answer in DNF, CNF, etc., which is identical to the form I've given here

Answer 8

@ganeshie8 be careful
S & (T | ~S) <=> T

Answer 9

correct me here :-

\( (\sim r \wedge p) \wedge (s \vee r) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

\( (\sim r \wedge p \wedge s) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

Answer 10

that is just he disjunctive syllogism I used

Answer 11

okay good so far

Answer 12

\( (\sim r \wedge p) \wedge (s \vee r) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

\( (\sim r \wedge p \wedge s) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

\( (\sim r \wedge p \wedge s) \wedge [(p \wedge t) \vee (\sim s \wedge \sim t) \vee (p \wedge \sim s)] \)

Answer 13

\( (\sim r \wedge p) \wedge (s \vee r) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

\( (\sim r \wedge p \wedge s) \wedge (\sim t \vee p) \wedge (t \vee \sim s) \)

\( (\sim r \wedge p \wedge s) \wedge [(p \wedge t) \vee (\sim s \wedge \sim t) \vee (p \wedge \sim s)] \)

\( (\sim r \wedge p \wedge s \wedge t) \)

Answer 14

oops ! ended up wid the same

Answer 15

$$\begin{align*}(\neg r\land p)\land(s\lor r)\land(\neg t\lor p)\land(t\lor\neg s)&\underbrace\Leftrightarrow_{D.S.} \neg r\land p\land s\land(\neg t\lor p)\land(t\lor\neg s)\\&\underbrace\Leftrightarrow_{D.E}\neg r\land p\land s\land (t\lor\neg s)\\&\underbrace\Leftrightarrow_{D.E.}\neg r\land p\land s\land t\end{align*}$$

Answer 16

OK, i think i got it, but can you just say me the other inference rule that you're using with DS.?