/jp/ - Otaku Culture

[Anonymous No.]

Okay, so I must be lacking something, I am proceeding on establishing that P=>(PvQ) is a tautology that the truth table is true in all cases, notwithstanding I am coming up with a row of false conclusions:

P | Q | (PvQ) | P=>(PvQ)
T | T | T | T
T | F | T | T
F | T | T | T
F | F | F | F

Is it just that I am missing some crucial point about how I should draw near this or am I just simply obtuse at math tonight? If someone could gracefully render me with some sort of approach to this problem or just a straight up suggestion as to what I am doing wrong I would be ever so grateful.

[Anonymous No.]

Nigga I can't understand a word of what you wrote

[Anonymous No.]

>>6406156

[Anonymous No.]

>>6406151
>I am proceeding on establishing that P=>(PvQ) is a tautology
>P=>(PvQ) is a tautology
.

[Anonymous No.]

If P and Q is false, then P=>Q is true.
Not /jp/ related

[Anonymous No.]

p => q is only false if p true and q false.
In your case you have

false => (false v false)
meaning:
false => false
and (false => false) has a logic value of true

[Anonymous No.]

P => (PvQ)
~P v (PvQ)
~P v P v Q
1 v Q
1 (tautology)

[Anonymous No.]

saten you baka

[Anonymous No.]

P => ( P & Q )

If dogs fly then dogs fly and OP also is an idiot.

(that's row 3 by the way, not 4 which you got wrong)

[Anonymous No.]

P => (P | v | Q )
T T T T T
T T T T F
F T F T T
F T F F F

The second column tells us that this is a tautology.

[sage No.]

Lookie: the same question was posted on yahoo answer, with slighty different wording.
(slighty different wording) => (same problem)
copypasta => ((slighty different wording) => (same problem))
true => (true => true)
true.
qed.

[Anonymous No.]

Why would you copy some random math question from Yahoo and post it here?