I broke logic

[Anonymous No.6380382]

Imagine a series circuit with two switches A and B and a lamp C.
If both switches are ON, then the lamp will be ON.

In propositional logic:
(A /\ B) => C

You can easily check the following logical entailment (go through all models of the LHS etc):
{(A /\ B) => C} |-- (A => C) \/ (B => C)

Therefore it is sufficient for the lamp to be ON if one of the switches in ON and the other is OFF.
This contradicts the physical design of a series circuit.

[Anonymous No.6380385]

>>6380382

Its not an OR operator, its an AND

[Anonymous No.6380387]

>>6380385

/\ means AND
\/ means OR

[Anonymous No.6380391]

>You can easily check the following logical entailment (go through all models of the LHS etc):
{(A /\ B) => C} |-- (A => C) \/ (B => C)

What? You're an idiot.

[Anonymous No.6380396]

>>6380391
Are you a CS major? Write down the truth table if you don't believe me. This should be known to everyone who took the first week of logic 101.

[Anonymous No.6380401]

>>6380396

You do realise that AnB => C means that you need both A=>C AND B=>C. Not one or the other. I can't even argue further because it's just so simple.

[Anonymous No.6380403]

>>6380401
What part of "logical entailment" do you not understand? Are you a CS major or something?

[Anonymous No.6380429]

>>6380401
http://www.wolframalpha.com/input/?i=truth+table+[%28A+and+B%29+%3D%3E+C]+%3C%3D%3E+[%28A+%3D%3E+C%29+or+%28B+%3D%3E+C%29]

[Anonymous No.6380449]

>>6380401
A logical entailment is in general not an equivalence.

Let X be a set of propositional expressions and Y be a propositional expression.
Then X logically entails Y ( in symbols: X |-- Y ) if each model of X is a model of Y.

Simple example:
{A /\ B} |-- A \/ B

Why? Because whenever A /\ B is true, then A is true and B is true. When A is true and B is true, then A \/ B is true.

Again: Are you a CS major? Why do you fail so epically at basic propositional logic?

What you posted is wrong btw.
>AnB => C means that you need both A=>C AND B=>C
Let A = true, B = false, C = false.

[Anonymous No.6380496]

OP here again.

A friend of mine is a psychology major and she found the error almost instantly. I'm still waiting for someone on /sci/ to have an IQ above room termperature.

When I come home from the gym and you faggots haven't figured it out, maybe I'm gonna give a hint.

[Anonymous No.6380576]

>>6380382
>(A /\ B) => C
Ok
>(A /\ B) => C} |-- (A => C) \/ (B => C)
I agree
>Therefore it is sufficient for the lamp to be ON if one of the switches in ON and the other is OFF.
What? That doesn't follow at all.

[Anonymous No.6380590]

>>6380576
This pretty much. I have no idea why you think it would mean that.

[Anonymous No.6380617]

>>6380382
>This contradicts the physical design of a series circuit.
Yes, because a physical circuit would follow the rule:
<span class="math">(A \land B) \Leftrightarrow C[/spoiler]

[Anonymous No.6380624]

>>6380449

OP, heres the thing, the essential argument form is that if the premises are true, the conclusion must also be true. It describes a valid argument form, not a valid argument.

So there is a hidden set of logical operators omitted in your premise, namely, that in a series circuit, the circuit is completed if and only if A and B are on. Therefore A or B alone is not sufficient to turn the circuit on.

You didn't break logic.

No one will ever break logic.

[Anonymous No.6380626]

>>6380624

All this is fairly clear without logical notation.

[Anonymous No.6380637]

>>6380449
-_-

you're really failing hard, dude.
I'm feeling bad for you.

see
>>6380576
{(A /\ B) => C} |-- (A => C) \/ (B => C)

is perfectly valid and the equivalence is ONE way to show that very easily.
Do you even model theory?


You seem to be the one epically failing here