Logic 101 (#40): Conditional Proofs

Conditional proofs allow you to assume that an antecedent is true, derive some consequent, and thereby demonstrate that the conditional statement is true.

[SlackwareNVM]

1. Pv(T-->U)
2. P-->Q
3. ~Sv~Q

4.     S         (The assumption)
5.     ~Q       (3, 4 disjunctive syllogism)
6.     ~P       (2, 5  MT)
7.     T-->U   (1, 6 disj. syll. again)
Therefore: S-->(T-->U)
As in "If we assume S is true, then T-->U must be true."

Thank you fore these videos, I hope to see some more.

[thetitanty]

Dude this actually helped me so much thanks

[tomding]

great job, my professor take an hour to explain this and ended up with the whole class having no idea what he is talking about

[blakeeaton]

4. S | Assumption for Conditional Proof
5. ~~S | Line 4 - Double Negation
6. ~Q | Lines 3 & 5 - Disjunctive Syllogism
7. ~P | Lines 2 & 6 - Modus Tollens
8. T => U | Lines 1 & 7 - Disjunctive Syllogism
9. S => (T => U) | Lines 4 - 8 - Conditional Proof

[fatimashirazi]

THANK YOU SO MUCH. MAY YOU GET ITS REWARD 🙌🏻

Thank you! overall great videos. watched them all. super useful for the paper that i'm currently writing

[gidrengidrenovi]

1. Pv(T->U)
2. P->Q
3. ~Sv~Q S->(T->U)
4. | S Assumption for conditional proof
5. | ~Q 3.4. Disjunctive syllogism
6. | ~Q-->~P 2. Contraposiiton
7. | ~P 5.6. Modus Ponens
8. | T->U 1.7. Disjunctive syllogism

[elisha]

I appreciate these videos but this system you're using is a lot simpler than what seems to be taught in some Logic classes that use a formal Fitch system.
As Fitch does not have syllogism rules. So what took you 8 steps actually requires about 18 when you actually derive the logic behind the syllogism rules

[jvgama]

If S is true, we can conclude ~Q (3).
If ~Q, we can conclude ~P (2).
If ~P, we can conclude T=>U (1).

Therefore, S => (T=>U)

[torosalvajebcn]

It could also be proven by using contraposition:



1) P v (T => U)

2)  P => Q

3) -S v -Q

4)      -( T => U)                  Assumption for conditional proof.

5)      1, 4, Disjunctive syllogism.

6)        2, 5 Modus ponens.

 3, 6 Disjuntiv syllogism.

8) -( T=> U) => -S               Conditional proof

9) S=>(T=>U)                     8 Replacement by contraposition.

[reniersteytler]

This was a bit difficult to grasp at the end. Are we just assuming that the conditional holds by virtue that we tested the assumption in line 4?

[TAEHSAEN]

Question: I thought you can only join S and T=>U through conjunction. Because they are both simultaneously true. So the answer should logically be S^(T=>U) by conjunction. How did you arrive at the conclusion that S=>(T=>U) because that is not the same as S^(T=>U)?

[Nay_]

i thought you only indented for indirect proof

[ThemisTheotokatos]

Great videos. Does this mean that by conditional proof, also S=>~P or ~Q=>P or ~Q=>T=>U? Does these mean that anything implies anything below it?

[DottMySaviour]

I'm trying to intuitive understand the concept of conditional proof. So, the idea is that when proofing P => Q, we are simply saying that we just want to proof that when P is true, Q must be true, while the other 3 possible combinations of P and Q doesn't matter. Am I right?

[cannos]

i pretty much did it the way shown in the video, except i put in the extra step of switching S to ~~S and then worked from there. but why dont i need to do that?

[CaptainASD]

Anybody actually solve this using fitch?

[AnjaliSharma-sgzg]

(∼A ⊃ B) v (A ⊃ R) ??? This statement solve it?

[paulcarello]

This proof appears to be incomplete.

1) P ∨ (T ⇒ U)
2) P ⇒ Q
3) ~S ∨ ~Q ∴ S ⇒ (T ⇒ U)

If we assume S is true, we can prove T -> U is also true (video proof)
∴ S ⇒ (T ⇒ U)

The question is, do we need to take an additional step and prove the conclusion when S is false
If S is false, then

S ⇒ (T ⇒ U) is vacuously true as S is false.