Determining Validity of Argument Using Indirect Truth Table

lifeiseasy
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The argument is:

~(A^B) / (~A^~B) -> (C^D) / D->C // C

I've attached the truth values that I've filled in.

I have difficulty in determining the truth values of A and B since I can't get any clue from other statements. Any advice would be appreciated.
 

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So perhaps you only need to know (or the problem can only determine) how many of A or B are true or false and not which one. Do you know that much, whether none, exactly one, at most one, at least one, or both need to be true or false?
 
Well it doesn't really matter. The question requires me to use an indirect truth table, so all premises are assumed to be true and conclusion false. If there is no contradiction in the truth values, that means all premises are true and the conclusion false, which implies that the argument is invalid. If at least one premise is false or the conclusion is true, that means the argument is valid. Now I'm stuck in the truth values of A and B, so I can't in turn determine the validity of the argument.
 
If there is *any* assignment of truth values to the atomic formulas that makes all premises true and the conclusion false, then the argument is invalid (i.e., not true under all assignments). Is there any such assignment? You only need one. What if A is true and B is false (and D and C are both false)?
 
"honestrosewater" is correct. The pivotal formula is (~A ^ ~B) -> (C ^ D) . Because C and D are false, and the value of the implication in this formula is set to "T", so must the value under the operator "^" (conjunct) be "F". Given the truth table for the conjunct, either or both of A and B can be F. Just so long as BOTH are not true, making the conjunct true, you are OK. All you need to do is find just ONE assignment of A and B to show the argument form to be invalid. Note that in many formal axiomatic presentations, upper-case letters A-M normally are reserved for instances (specific situations - substitution instances) or abbreviations for actual statements. The letters p-z often are reserved for variables. Also, when you mix set theory with logical formalisms, the capital letters designate sets, although ^ and v are intersection, and union of sets, respectively. Just be aware of your context.
 
They cannot both be false.

Since ~(A^B) must be true, the conjunction must be false, so at least one of A or B must be false.

Since (~A^~B) must be false, at least one of the conjuncts must be false, which means at least one of A or B must be true.

So at least one is true and at least one is false, and since there are only two of them, exactly one is true and exactly one is false. It doesn't matter which one is which since conjunction is commutative.
 
For he whole expression (~A^~B) -> (C^D), the component (~A ^ ~B) cannot be true, because -> is assigned a T and the C^D is false. This means the antecedent CANNOT be a T if the truth of the material implication is to be preserved. For the original questioner, recall:

p q p -> q
- - - -- -
F F T
F T T
T F F
T T T

The only assignment lines (under the p and q columns) lines where the consequent, q, is assigned a false are the first and the third, F -> F and T -> F. Obviously, you don't want the T -> F, if you want to preserve the truth value of the implication. Hence, there must be a F under the conjunct, ^, in ~A ^ ~B. This means that either the whole expression ~A or ~B can be T, but both can NOT be true, given the truth table for the conjunct:

p q p ^ q
- - - -- -
F F F
F T F
T F F
T T T

That is, both A and B, themselves cannot be false, as honestrosewater says, as we see in the truth table:

~A ^ ~B
--- -- ---
T F T T F

The unary operator, ~, is the operator under which truth values for determining the truth value for the conjunct are placed.

To recapitulate, "They cannot both be false. " - refers to A and B, as the T's are under the ~. I was referring to BOTH ~A and ~B being true (making the whole expression ~A ^ ~B true), and I should have been more explicit. Instead of "Just so long as BOTH are not true", I should have said, "Just so long as BOTH ~A and ~B are not true...".

What honestrosewater says about ~(A^B) being true and the inner expression being false, with EITHER A or B being true is, of course, true. Make sure, however, that consistent assignments of T and F for A and B are made.

These small windows are very hard to write in.
 
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