I'm lost on b, too. This is my best guess: If you know A is true and that A=>B, then you know that B must be true because the only other option is for A to be true and A=>B to be false meaning that B is also false.
A | B | A=>B| A^(A=>B)| A^(A=>B)=>B T | T | T | T | T T | F | F | F | T F | T | T | F | T F | F | T | F | T
When I read the statement "if you know A and you know A=>B, then you can conclude B", I read it as "if you know the 1st and 3rd columns, you can figure out what B is or what is the 2nd column.
But when we look at the last two rows, B is different but the 1st and 3rd column are the same. Am I missing something?
I think it's a modus polens (Latin for "the way that affirms by affirming") because it's a "tautology" or "logical validity," i.e., all combinations of T and F in the antecedents yield a true consequent. So, if you can get to this as a point in your proof, everything becomes true as a consequence [we should be so lucky in our daily lives ;-)].
I'm pretty sure that's what Max said, but maybe with fewer words.
"When I read the statement "if you know A and you know A=>B, then you can conclude B", I read it as "if you know the 1st and 3rd columns, you can figure out what B is or what is the 2nd column."
I don't think that's what it's saying. I think it's more like, "if you know that 'A' is true, and you know that 'A=>B' is true, then you can find the value for 'B'.
So to see that in the truth table, look at the column for "A^(A=>B)" - it is only true when both "A" and "A=>B" are true. In this case, you can see that "B" is also true. For all other values, it is not true... in other words, "if you know at "A" is true, and you know that "A=>B" is true, then you can deduce that "B" is also true.
The column for "A^(A=>B)=>B" demonstrates for for ALL values of "A" and "B," the statement "A^(A=>B)=>B" IS true (even if the specific values of "A" and "B" are not), which demonstrates the principle of modus polens.
But I do see the potential for confusion here. "If you know A and you know A=>B" does make it sound like, "if you know the values of these things" rather than, "if you know that A is true and A=>B is true."
I think that's part of what we're supposed to deduce: - the "A^(A=>B)" column only holds if A and A=>B are both true, and... - the "A^(A=>B)=>B" column is true for all values of A and B, therefore we can trust in the principle.
Well, here's my truth table... I'll let someone else tackle part b. :)
ReplyDeleteΦ | Ψ | Φ ⇒ Ψ | Φ ⋀ (Φ ⇒ Ψ) | [Φ ⋀ (Φ ⇒ Ψ)] ⇒ Ψ
T | T | T | T | T
T | F | F | F | T
F | T | T | F | T
F | F | T | F | T
I got the same truth table but do not understand how this demonstrates modus pone s.
ReplyDeleteI'm lost on b, too. This is my best guess: If you know A is true and that A=>B, then you know that B must be true because the only other option is for A to be true and A=>B to be false meaning that B is also false.
ReplyDeleteHere's my take on part b:
ReplyDeleteif you know Φ and you know Φ ⇒ Ψ, then you can conclude Ψ.
In other words, if you know that something and that something implies something else, then you know the something else.
For example: "I can see stars." and "if you can see stars, the sun has set." Therefore, in this case, we can determine that the sun has set.
The truth table shows that Φ ⋀ (Φ ⇒ Ψ) is true iff Φ, Ψ are both true.
It also shows that for all values of Φ, Ψ, "[Φ ⋀ (Φ ⇒ Ψ)] ⇒ Ψ" is true.
Thoughts?
*and therefore, demonstrates that modus ponens is a valid rule of
ReplyDeleteinference.
Can someone help me with this if A is false?
ReplyDeleteA | B | A=>B| A^(A=>B)| A^(A=>B)=>B
T | T | T | T | T
T | F | F | F | T
F | T | T | F | T
F | F | T | F | T
When I read the statement "if you know A and you know A=>B, then you can conclude B", I read it as "if you know the 1st and 3rd columns, you can figure out what B is or what is the 2nd column.
But when we look at the last two rows, B is different but the 1st and 3rd column are the same. Am I missing something?
Thanks for any assistance.
I think it's a modus polens (Latin for "the way that affirms by affirming") because it's a "tautology" or "logical validity," i.e., all combinations of T and F in the antecedents yield a true consequent. So, if you can get to this as a point in your proof, everything becomes true as a consequence [we should be so lucky in our daily lives ;-)].
ReplyDeleteI'm pretty sure that's what Max said, but maybe with fewer words.
Hey Molly -
ReplyDelete"When I read the statement "if you know A and you know A=>B, then you can conclude B", I read it as "if you know the 1st and 3rd columns, you can figure out what B is or what is the 2nd column."
I don't think that's what it's saying. I think it's more like, "if you know that 'A' is true, and you know that 'A=>B' is true, then you can find the value for 'B'.
So to see that in the truth table, look at the column for "A^(A=>B)" - it is only true when both "A" and "A=>B" are true. In this case, you can see that "B" is also true. For all other values, it is not true... in other words, "if you know at "A" is true, and you know that "A=>B" is true, then you can deduce that "B" is also true.
The column for "A^(A=>B)=>B" demonstrates for for ALL values of "A" and "B," the statement "A^(A=>B)=>B" IS true (even if the specific values of "A" and "B" are not), which demonstrates the principle of modus polens.
Make sense?
But I do see the potential for confusion here. "If you know A and you know A=>B" does make it sound like, "if you know the values of these things" rather than, "if you know that A is true and A=>B is true."
DeleteI think that's part of what we're supposed to deduce:
- the "A^(A=>B)" column only holds if A and A=>B are both true, and...
- the "A^(A=>B)=>B" column is true for all values of A and B, therefore we can trust in the principle.