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.
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
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 ))
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 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.
