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I am trying to prove a statement of the form:

If A and B, then C.

Is this equivalent to the following statement?

Given A, if B, then C.

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Nope. If F (false) then F (or T (for true)) is always true. So in the first statement if A is F then it is always true whatever values of B or C. But this case is not covered in the second statement where A is always T. Clear ? –  brunoh Oct 17 '11 at 17:02
    
It's hard to tell what you mean. Does "If A and B, then C." mean "((A^B)->C))" or "(A->(B->C))"? Both of those come as logically equivalent. Does "Given A, if B, then C." mean (A->(B->C)) or does it mean A|=(B->C), where the double turnstile "|=" can get read as "entails"? You can convert these into each other via completeness, the deduction theorem and its converse, and soundness, so in some sense they come as equivalent. But, since one of them comes as happening in the object language (the first one), and the second one in the metalanguage, in some sense they aren't equivalent. –  Doug Spoonwood Oct 17 '11 at 17:27
    
If "Given A, if B, then C." means A|=(B->C), or A|-(B->C) [A proves or yields (B->C)], and "If A and B, then C." means either of the above, in some sense the two statements come as equivalent, in some sense they do not come as equivalent since one happens in the metalanguage and the other in the object language. –  Doug Spoonwood Oct 17 '11 at 17:29

2 Answers 2

up vote 6 down vote accepted

I will interpret your question as asking whether the two statements $(A\land B)\implies C$ and $A\implies(B\implies C)$ are logically equivalent. They are.

To verify this, we can proceed slowly, by calculating the truth-table for each. Not too bad, only $8$ entries for each table. We can cut down the work by noting that if $A$ is false, each of our two sentences is easily seen to be true.

Or else we can proceed "algebraically." The sentence $(A\land B)\implies C$ is logically equivalent to $\lnot(A\land B) \lor C$, which in turn is equivalent to $(\lnot A \lor \lnot B)\lor C$.

The sentence $A\implies(B\implies C)$ is logically equivalent to $\lnot A \lor(B\implies C)$, which in turn is equivalent to $\lnot A \lor (\lnot B \lor C)$. Now we are essentially finished.

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You can also cut down the work, but noting that if (1^x)==x, and (1->x)==x, where "==" indicates logical equivalence. So, if "a" is true, then we have ((1^b)->c))==(b->c) and (1->(b->c))==(b->c) for both statements, so they come as equivalent. –  Doug Spoonwood Oct 17 '11 at 17:21

$(A \wedge B) \Rightarrow C$ means that when $A$ and $B$ are both true, so is $C$. No surprises here...

But your statement, $A \wedge (B \Rightarrow C)$ is different, since the conditional $B \Rightarrow C$ can be true when $B$ is false.

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I would say that "Given A, if B then C" means what André said, not what The said. –  GEdgar Oct 17 '11 at 17:17

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