0:46 Inference Rules 1:09 A & B are True • A is True • B is True Not not & Not and (not both) Or ( a or b) 2:45 Conditional 4:22 Modus Ponens 4:49 If P, Then Q P Not Q Assume Not P Assume Q Uptack = close the branch Closed Tree Branches Close with 2 Contradictions on the _same_ branch Apply every possible rule 7:55 Assume the negation of the conclusion *Logical Truth Tautology* 8:22 P v Q
The only time a conditional is false is when the antecedent is true and the consequent is false. In every other case, it's true. So yes, if A and B are true, A->B is true. But A->B is also true when both A and B are false, and when A is false and B is true.
To be more clear than the other reply, the branches (or lack of branches e.g. for a conjunction) have the *sufficient* conditions for the sentence that is being expanded. B on its own is sufficient for A->B, as is ~A. This is exactly equivalent to replacing A->B with its equivalent, (~A)V(B), and decomposing that disjunction, which of course gets you the two aforementioned single-atom branches.
Also, incidentally, if you connect two statements that you think are the same with a then it will end up being a tautology, if they are equivalent because a double implication is only wrong if the two statements' truth values are different under one interpretation. E.g.: (A B) (B A)
***** Right, it's only a tautology if both claims on either side of the double implication are logically equivalent, otherwise it is a contingent or a contradiction and, therefore, logically inequivalent. Your example of (A B) (A C) evaluates as false when A = false, B = false and C = true, so not a tautology nor are the two claims either side of it equivalent to each other.
I tryied the truth trees on some tautologies but non of my branches close. Even though I prove the tautology using truth tables. My expresion is like this: ~(p & q) => ((p OR r) => ( p =>r ))
You lost me at 5:25. This is a common thing with many teachers. They don't explain what they're teaching properly which is why students find it so hard to understand.
I dont find that youre actually setting up any methodology that makes sense, you just talking in terminology that confuses When you say -(A and B), and draw two arrows that lead to false A and False b, what have you actually done? I dont know, maybe you said that a could be false in a case where b was also false, but what exactly does that mean? Sorry im venting a little bit over the common teaching style
If you’re referring to 1:30 … ~ is a negation operator. Negation switches truth value of a set/variable. If ~ is in the scope of (A & B), it switches the truth value of the set within the bracket (given the basic function of a bracket). This is to assume the values were true without the negation. Thereby, ~ then renders the entire (A&B) false. When dissecting the set ~(A&B), it logically reveals that both ~A and ~B are false in this manner, and this is, as mentioned, given that ~ is a negation operator and the basic functions of a bracket. (Again, this is assuming that the A and B values were true before the negation.) I think his explanation were well and fair.
0:46 Inference Rules
1:09 A & B are True
• A is True
• B is True
Not not
&
Not and (not both)
Or ( a or b)
2:45 Conditional
4:22 Modus Ponens
4:49
If P, Then Q
P
Not Q
Assume Not P
Assume Q
Uptack = close the branch
Closed Tree
Branches Close with 2 Contradictions on the _same_ branch
Apply every possible rule
7:55 Assume the negation of the conclusion
*Logical Truth Tautology*
8:22 P v Q
The only time a conditional is false is when the antecedent is true and the consequent is false. In every other case, it's true. So yes, if A and B are true, A->B is true. But A->B is also true when both A and B are false, and when A is false and B is true.
Basic
Thanks, this helped a lot. Now teach me your accent please.
To be more clear than the other reply, the branches (or lack of branches e.g. for a conjunction) have the *sufficient* conditions for the sentence that is being expanded. B on its own is sufficient for A->B, as is ~A. This is exactly equivalent to replacing A->B with its equivalent, (~A)V(B), and decomposing that disjunction, which of course gets you the two aforementioned single-atom branches.
thank you SO much you have no idea
Thank you, I had no idea how they worked before watching this video
There is something I don't quite grasp:
Is A ↔ B the same as B ↔ A?
+RnBandCrunk yes
it's equivalent to (A -> B) && (B -> A)
They are logically equivalent.
Also, incidentally, if you connect two statements that you think are the same with a then it will end up being a tautology, if they are equivalent because a double implication is only wrong if the two statements' truth values are different under one interpretation.
E.g.: (A B) (B A)
e.g. If (A B) (A C) or similar, then it wouldn't necessarily mean that it is an tautology, is that what I'm understanding here?
*****
Right, it's only a tautology if both claims on either side of the double implication are logically equivalent, otherwise it is a contingent or a contradiction and, therefore, logically inequivalent.
Your example of (A B) (A C) evaluates as false when A = false, B = false and C = true, so not a tautology nor are the two claims either side of it equivalent to each other.
is this Gareth
I tryied the truth trees on some tautologies but non of my branches close. Even though I prove the tautology using truth tables. My expresion is like this: ~(p & q) => ((p OR r) => ( p =>r ))
Thank you!
Enjoyed this thank you!
You lost me at 5:25. This is a common thing with many teachers. They don't explain what they're teaching properly which is why students find it so hard to understand.
thank you my friend!
Thank you for this!
Beautiful!
Thank you
Thx!
I dont find that youre actually setting up any methodology that makes sense, you just talking in terminology that confuses
When you say -(A and B), and draw two arrows that lead to false A and False b, what have you actually done? I dont know, maybe you said that a could be false in a case where b was also false, but what exactly does that mean?
Sorry im venting a little bit over the common teaching style
If you’re referring to 1:30 …
~ is a negation operator. Negation switches truth value of a set/variable.
If ~ is in the scope of (A & B), it switches the truth value of the set within the bracket (given the basic function of a bracket).
This is to assume the values were true without the negation. Thereby, ~ then renders the entire (A&B) false.
When dissecting the set ~(A&B), it logically reveals that both ~A and ~B are false in this manner, and this is, as mentioned, given that ~ is a negation operator and the basic functions of a bracket.
(Again, this is assuming that the A and B values were true before the negation.)
I think his explanation were well and fair.
thanks a lot!
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