Question

Evaluate: \[[\neg p \wedge (p \vee q)] \rightarrow q\]

Answer 1

i suppse \[\neg p \wedge (p \vee q) \equiv p\]

Answer 2

so then it becomes \[p \rightarrow q\]

Answer 3

then what?

Answer 4

oh i see

Answer 5

i made a mistake

Answer 6

\[\neg p \wedge (p \vee q) \equiv F \vee (\neg p \wedge q) \rightarrow q\]

Answer 7

then this becomes \[T \wedge (p \vee q) \vee q\]

Answer 8

then \[T \wedge (p \vee q) \equiv p \vee q \]

Answer 9

in my solution i will be treating them the same....my latex is gonna get confusing if i follow the rules

Answer 10

\[\equiv p \vee q \vee q\]
\[q \vee q \equiv T\]
so.. \[\equiv p \vee T \equiv T\]
yes?

Answer 11

so then this is a tautology?

Answer 12

implication law is
P->Q=nP V Q
Right?
Sorry for no Latex,

Answer 13

yes

Answer 14

so is that a yes to my question?

Answer 15

yes it's a tautology, but you messed something up and somehow ended up with a tautology anyways lol

Answer 16

where?

Answer 17

\[\small¬p∧(p∨q)\Rightarrow q\iff(\neg p\wedge p) \vee(\neg p\lor q)\Rightarrow q\iff (\neg p\lor q)\Rightarrow q \iff (\neg p\Rightarrow q)\vee(q\Rightarrow q )\]\[\iff q\Rightarrow q\qquad \top\]

Answer 18

hmm looks different...

Answer 19

implication is distributive?

Answer 20

seems you're the one who went wrong @PhoenixFire ....

Answer 21

\[(\neg p \wedge q) \rightarrow q\]
should become \[\neg(\neg p \wedge q) \vee q\]
by DM law
\[p \vee q \vee q\]

Answer 22

hmm seems i missed a step too

Answer 23

nevertheless, important thing is.. i was right....that was my question in the first place anyways

Answer 24

de morgans law.... you have to negate both.
n(nP ^ Q) V Q
becomes
nnP V nQ V Q

Answer 25

nnP <-- Involution law becomes P.
What you missed was distributing the negative to the Q during De Morgan's Law