¬(Φ ⋁ Ψ) and (¬Φ) ⋀ (¬Ψ) are equivalent 1. Φ ⋁ Ψ means that at least one of Φ and Ψ are true 2. Thus ¬(Φ ⋁ Ψ) means that it is not the case that at least one of Φ and Ψ is true. 3. If it is not the case that at least one of Φ and Ψ are true, then they are both false. 4. This is clearly the same thing as saying that both ¬Φ and ¬Ψ are true. 5. By the meaning of and, this can be expressed as (¬Φ) ⋀ (¬Ψ).
1. Φ ⋁ Ψ means not both Φ and Ψ are false. 2. Thus, ¬(Φ ⋁ Ψ) means that it is not the case that both Φ and Ψ are false. 3. If not both Φ and Ψ are false, then by negation both Φ and Ψ are true. 4. By the meaning of "and," ¬Φ and ¬Ψ is true because both both Φ and Ψ are true. 5. Therefore, ¬(Φ ⋁ Ψ) and (¬Φ) ⋀ (¬Ψ) are equivalent.
I'm not sure this is a valid proof, however. I like Max's better. What do others think?
Note that in your step three, you have "both Φ and Ψ are true" but in step four you state, " ¬Φ and ¬Ψ is true because both both Φ and Ψ are true."
In order for "¬Φ and ¬Ψ" to be true, both Φ and Ψ must be false.
I'm ashamed to admit it, but I spent nearly an hour to come up with what I did. It's so obvious on the face of it, but getting it all laid out was trickier than I thought it would be.
¬(Φ ⋁ Ψ) and (¬Φ) ⋀ (¬Ψ) are equivalent
ReplyDelete1. Φ ⋁ Ψ means that at least one of Φ and Ψ are true
2. Thus ¬(Φ ⋁ Ψ) means that it is not the case that at least one of Φ and Ψ is true.
3. If it is not the case that at least one of Φ and Ψ are true, then they are both false.
4. This is clearly the same thing as saying that both ¬Φ and ¬Ψ are true.
5. By the meaning of and, this can be expressed as (¬Φ) ⋀ (¬Ψ).
Looks good to me. According to grading rubric you also need a concluding statement.
ReplyDeleteGood point, Susan! I hadn't attempted the problem set when I did this, and hadn't made the connection. Thanks!
ReplyDeleteI started working on this from the point of:
ReplyDelete1. Φ ⋁ Ψ means not both Φ and Ψ are false.
2. Thus, ¬(Φ ⋁ Ψ) means that it is not the case that both Φ and Ψ are false.
3. If not both Φ and Ψ are false, then by negation both Φ and Ψ are true.
4. By the meaning of "and," ¬Φ and ¬Ψ is true because both both Φ and Ψ are true.
5. Therefore, ¬(Φ ⋁ Ψ) and (¬Φ) ⋀ (¬Ψ) are equivalent.
I'm not sure this is a valid proof, however. I like Max's better. What do others think?
Oops! Maybe Step 4 should be, "... because both ¬Φ and ¬Ψ (step 4) are true."
ReplyDeleteSee why I don't like it? :-p
Can anyone make it clearer starting with Φ and Ψ both false?
I started that way, too, and abandoned it.
ReplyDeleteNote that in your step three, you have "both Φ and Ψ are true" but in step four you state, " ¬Φ and ¬Ψ is true because both both Φ and Ψ are true."
In order for "¬Φ and ¬Ψ" to be true, both Φ and Ψ must be false.
I'm ashamed to admit it, but I spent nearly an hour to come up with what I did. It's so obvious on the face of it, but getting it all laid out was trickier than I thought it would be.
No wonder I couldn't make any headway. I had my own tautology.
DeleteI spent a lot of time on it, too. I remember doing these as a sophomore in high school math, but clearly I didn't remember HOW to do them.
Thanks, Max.