Derive: $(A \implies (B \wedge C)) \models (A\implies B)$

I need help!

I am taking a Math & Truth Course and there are logic and paradox problems on an assignment I don't understand.

Anyone willing to help me derive the following?

$$(A \implies (B \wedge C)) \models (A\implies B)$$

Note: In the above equation the & sign between B & C is actually an upside down "u" in the problem however I am unable to locate a way to present an upside down "u" which means "and" however I have read that you are to place & sign there on the computer. And the "F" is in place of what appears to be a line with two lines directly from it...

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Are you trying to write $(A \implies (B \cap C)) \vDash (A \implies B)$? If so, $\vDash$ may mean semantic consequence and you have probably been taught how to manipulate it – Henry Dec 3 '11 at 0:15
Can you check if my latex edit is what you meant? – user13838 Dec 3 '11 at 0:25
So, according to the comments/answers: you should discuss this with the course instructor, because here we do not know your deductive system... – GEdgar Dec 3 '11 at 0:49
By now long-standing web tradition, all caps is considered shouting. (see the first paragraph of the Wikipedia article on "all caps"). Use them sparingly. – Arturo Magidin Dec 3 '11 at 5:09

3 Answers

If $A$ implies both $B$ and $C$, then in particular, $A$ implies $B$.

Thus, in any context where $A\to(B\wedge C)$ is true, then also $A\to B$ is true. So $A\to(B\wedge C)\models A\to B$.

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But your title makes me think that perhaps you are not asking about the validity $A\to(B\wedge C)\models A\to B$, but rather asking for a derivation, that is, establishing $A\to(B\wedge C)\vdash A\to B$, and the answer to this will depend very much on the details of the formal proof system that you are using. There are many. – JDH Dec 3 '11 at 0:45
I am to use logical implications ... some of the rules would be premise, repetition, assumption, -> introduction, -> elimination, (upside down v ) introduction, (upside down V) elimination, v introduction, v elimination, (arrows pointing both ways) introduction, (arrows pointing both ways) elimination, not introduction and not elimination are the list of rules to apply. The instructor says this problem is direct and should only use 4 statements. Does this help any? – Amy Hinkley Dec 11 '11 at 23:54

You might write a truth table.

You might also note that if A is false, then (A⟹B) is true also. Then suppose A true, and (A⟹(B∧C)) true also. It then follows that (B∧C) is true also. So, B holds true. Thus, (A⟹B) holds true additionally. So, in either case (A⟹(B∧C))⊨(A⟹B)

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I am to use logical implications ... some of the rules would be premise, repetition, assumption, -> introduction, -> elimination, (upside down v ) introduction, (upside down V) elimination, v introduction, v elimination, (arrows pointing both ways) introduction, (arrows pointing both ways) elimination, not introduction and not elimination are the list of rules to apply. The instructor says this problem is direct and should only use 4 statements. Does this help any? – Amy Hinkley Dec 11 '11 at 23:55

From your comment you want to show that (A⟹(B∧C))|-(A⟹B) instead of that "(A⟹(B∧C))|=(A⟹B)". The difference comes as that "|-" means that the left side "yields" or "proves" the right. If you have "|=", then the left side makes the right side "valid" or "entails" it. You're using what gets called a "natural deduction" system.

Alright, now note that (A⟹B) consists of a conditional statement. With conditional statements, a good general method comes as to assume the antecedent (the part on the left side of the "⟹"), derive the consequent (the part on the right), and finally use conditional (⟹) introduction. So, here you would assume A, and hope to derive B somehow, and then use conditional introduction to get (A⟹B). You also have (A⟹(B∧C)) and A what can you can you infer from your introduction and elimination rules? Once you have that, what else can you infer from what you just got along with your rules?

Does that help?

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Does this make sense? A implies B and C premise, A as an assumption, B and C -> elimination, C (upside down V) elimination, A v introduction. A implies B -> introduction ??? – Amy Hinkley Dec 14 '11 at 5:17
@AmyHinkley This makes sense "A implies B and C premise, A as an assumption, B and C -> elimination, C (upside down V) elimination,". This doesn't "A v introduction.". I think you're close. – Doug Spoonwood Dec 14 '11 at 13:08
Please tell me what would make sense? Would I introduce B verses A? And then what? I am stumped and just want to ease my mind. Thank you for your help thus far. – Amy Hinkley Dec 15 '11 at 4:35
1 (A⟹(B∧C)) premise 2 A assumption 3 (B^C) 1, 2 -> elimination 4 B ^ elimination 5 (A⟹B) 2-4 ⟹ introduction. The only thing missing there comes to keep track of scope. – Doug Spoonwood Dec 15 '11 at 12:18
Wonderful ... my mind is at ease now ... Curious, the logical operations (=> and <->) correspond to relations on sets, can you identify these? There are two. As well, what are set theory expressions for A => B, A => (B(and, not C)? And is there anything in logic that could not be expressed in terms of set theory and vice versa? – Amy Hinkley Dec 15 '11 at 20:33