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
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.
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.
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)?
+TAEHSAEN I'm not sure if there's a better answer than "that's just how conditional proofs work." When you begin a conditional proof, you choose something to assume. Let's call that A. Then you can apply the rules of logical inference as normal. After, the rules of conditional proofs say that you may conclude that A => [ ], where [ ] is whatever you found through the standard proof process. The idea is that the first line assumes that something is true, and the rest of the proof techniques show that other things must be true CONDITIONAL on the assumption that you made initially. Hence you use the conditional operator =>.
+William Spaniel Alright thanks. I really appreciate the hard work you put into this. I'm passing my logic class because of you :) You're a super smart and dedicated dude :)
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?
The way that I understood it is that by proving that when P is true Q must also be true, is enough to prove the veracity of the conditional statement P=>Q. Because if it is actually true that when P is true Q is also true, it is also true that when P is true Q can't be false so the second possible combination is already proven as not possible by the conditional proof. Then the other two possible combinations would be with P being false (and Q being either true or false). No matter what, if any of those two last combinations were true, still the conditional statement P=>Q would remain vacuously true. So once we've ensured that when P is true Q follows, then the conditional statement is proved no matter if in reality P turns to be true or false.
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) P 1,4, Disjunctive syllogism. 6) Q 2,5 Modus ponens. 7) -S 3,6 Disjuntiv syllogism. 8) -( T=> U) => -S Conditional proof 9) S=>(T=>U) 8 Replacement by contraposition.
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?
+Nicholas Haley It is a matter of taste. One adherence to disjunctive syllogism is that it there must be the negation of exactly what appears on the left side of the disjunction. (This is what you did.) But if you get out a truth table, disjunctive syllogism works the same if the left side of the disjunction is a negation and you have a line with the non-negated left side of the disjunction. (This is what's in the video.) If you are taking a class, check with your professor to see what is allowed for the purposes of grading.
@@PunmasterSTP This was so long ago. Haha but tbh I got a C in logic. To make up for it I went to Law School. My point being that one grade, class, or not so successful day wont stop you from your dreams. Best.
=> is also written as つ
Dude this actually helped me so much thanks
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
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
Thank you! overall great videos. watched them all. super useful for the paper that i'm currently writing
How did your paper turn out?
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.
THANK YOU SO MUCH. MAY YOU GET ITS REWARD 🙌🏻
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.
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
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)
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)?
+TAEHSAEN I'm not sure if there's a better answer than "that's just how conditional proofs work." When you begin a conditional proof, you choose something to assume. Let's call that A. Then you can apply the rules of logical inference as normal. After, the rules of conditional proofs say that you may conclude that A => [ ], where [ ] is whatever you found through the standard proof process.
The idea is that the first line assumes that something is true, and the rest of the proof techniques show that other things must be true CONDITIONAL on the assumption that you made initially. Hence you use the conditional operator =>.
+William Spaniel Alright thanks. I really appreciate the hard work you put into this. I'm passing my logic class because of you :) You're a super smart and dedicated dude :)
Anybody actually solve this using fitch?
i thought you only indented for indirect proof
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?
The way that I understood it is that by proving that when P is true Q must also be true, is enough to prove the veracity of the conditional statement P=>Q. Because if it is actually true that when P is true Q is also true, it is also true that when P is true Q can't be false so the second possible combination is already proven as not possible by the conditional proof. Then the other two possible combinations would be with P being false (and Q being either true or false). No matter what, if any of those two last combinations were true, still the conditional statement P=>Q would remain vacuously true. So once we've ensured that when P is true Q follows, then the conditional statement is proved no matter if in reality P turns to be true or false.
I think you are correct, and that is a great way to sum up the main idea of conditional proofs!
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?
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
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?
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) P 1,4, Disjunctive syllogism.
6) Q 2,5 Modus ponens.
7) -S 3,6 Disjuntiv syllogism.
8) -( T=> U) => -S Conditional proof
9) S=>(T=>U) 8 Replacement by contraposition.
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?
+Nicholas Haley It is a matter of taste. One adherence to disjunctive syllogism is that it there must be the negation of exactly what appears on the left side of the disjunction. (This is what you did.) But if you get out a truth table, disjunctive syllogism works the same if the left side of the disjunction is a negation and you have a line with the non-negated left side of the disjunction. (This is what's in the video.) If you are taking a class, check with your professor to see what is allowed for the purposes of grading.
+William Spaniel thank you for the explinantion! i helive my prof. prefers i show the double negation.
@@cannos09 Hey, how did the rest of your class go?
@@PunmasterSTP This was so long ago. Haha but tbh I got a C in logic. To make up for it I went to Law School. My point being that one grade, class, or not so successful day wont stop you from your dreams. Best.
@@cannos09 It was great to hear from you, and I think that is an amazing outlook! How is law school going, or did you already graduate?
brilliant
(∼A ⊃ B) v (A ⊃ R) ??? This statement solve it?