If B then C, thus if (A and B) then C

>If B then C, thus if (A and B) then C

Why the fuck can;t I wrap my mind around this? Can anyone do a proof?

(A and B) then B
B then C

How is (A and B) then B logical?

Maybe I'm just dumb but let me give this a shot:

(A and B) is the intersection between A and B so it is a subset of both A and B. So anything in (A and B) is also in B. So we can say if B then C, thus if subset(B) then C as an equivalent expression

If we know both A and B are true statements, then we know (trivially) that B is a true statement.
If we know Adam and Bill are humans, then we know Bill is a human.

You made that way more confusing than it needed to be.

>>If B then C, thus if (A and B) then C
How about this:
A = has brown hair
B = is human
C = has two legs

(A and B) describes an object which is both human and has brown hair. Therefore said object has two legs

I think you got this OP but

A AND B means A = 1, B =1

B =1 then C = 1

thus

A and B then C

And vs or is the key kohai

If B is a true statement, then C is a true statement.

A's truthness is unrelated to C's. So if A = true AND B = true, we know that C = is true because the statement that matters (B) is true.

(A and B) true means that A is true and B is true. If B is true, then C is true. Thus, if (A and B) is true, then A is true and B is true, and if B is true, then C is true.

Conjunction elimination

[math]
\displaystyle
\begin{matrix}
\underline{A} & \underline{B} & \underline{C} & \underline{B \rightarrow C}& \underline{B \cdot C \rightarrow C} \\
0 & 0 & 0 & \text{0 [1] 0} & \text{0(0)0 [1] 0} \\
0 & 0 & 1 & \text{0 [1] 1} & \text{0(0)1 [1] 1} \\
0 & 1 & 0 & \text{1 [0] 0} & \text{1(0)0 [1] 0} \\
0 & 1 & 1 & \text{1 [1] 1} & \text{1(1)1 [1] 1} \\
1 & 0 & 0 & \text{0 [1] 0} & \text{0(0)0 [1] 0} \\
1 & 0 & 1 & \text{0 [1] 1} & \text{0(0)1 [1] 1} \\
1 & 1 & 0 & \text{1 [0] 0} & \text{1(0)0 [1] 0} \\
1 & 1 & 1 & \text{1 [1] 1} & \text{1(1)1 [1] 1}
\end{matrix}
[/math]

That must have taken some serious autism to type out.

not really, 90% of it is just copy+paste

A, B, and C are statements, not sets.

Given: B -> C

Suppose (A & B)
then B is true
then C follows from (B -> C)
Thus (A & B) -> C

Mate, imagine A and B are light switches. If you flick the B switch, light bulb C will turn on.

Light switch A doesn't have to be connected to anything.

So, if you flick both A and B, light bulb C turns on.

this.
what is this, kindergarten logic 101?

fucking disgusting

given: B -> C
prove: (A and B) -> C
A and B
B
C (Modus ponens)
(A and B) -> C
QED

Proving an implication involves assuming the antecedent and showing that it logically leads to the consequent.

What if A and C are mutually then D, where D and C are mutually exclusive?

This is logically correct. Do the truth table.

This, in my logic class this was day one and everyone understood because they aren't fucking mongoliods.