Thread: A Combination of True and False Value unspecified for Implication Logical Operator

Attached is a PDF document I found while looking for an article on the If-then/Implication logical operator. The truth table for If/Then is on the 8th page of the PDF file.


What if it is known that the implication compound statement is False and it is known that the antecedent is False? What can one conclude (if one can at all) about the consequent since the truth table doesn't specify the assumption that the implication compound statement is false?

Couldn't we just treat the implication compound statement as a statement and apply the implication operator and calculate it according to to the implication truth table? By this I mean: Let A,B be defined mathematical statements and C be an undefined math statement.

A is False
A=>B is False
Let A=>B be C. So C is False.

So we are essentially evaluating A=>C. Both A and C are false so according to the truth table, the statement A=>C is True.

But then I wouldn't be answering my question ... since I'm essentially computing another implication problem (namely the implication of A and C) instead of deducing what the consequent might be. Hmm.

So the question remains: What if it isn't defined by the truth table?

Originally Posted by Elusive1324

What if it is known that the implication compound statement is False and it is known that the antecedent is False?

This is impossible under the standard definition, which is in the PDF document.

Originally Posted by Elusive1324

So the question remains: What if it isn't defined by the truth table?

What is "it"?

Originally Posted by emakarov

What is "it"?

The combination of constituent statements A and B and their implication when A and the implication are both false.

Originally Posted by emakarov

This is impossible under the standard definition, which is in the PDF document.

Do you mean impossible as in the deduction will not constrain the truth value of B because it is undefined? Or did you mean impossible as in we'll never encounter that situation?

Originally Posted by Elusive1324

What if it is known that the implication compound statement is False and it is known that the antecedent is False?

Originally Posted by emakarov

This is impossible under the standard definition, which is in the PDF document.

Originally Posted by Elusive1324

Do you mean impossible as in the deduction will not constrain the truth value of B because it is undefined? Or did you mean impossible as in we'll never encounter that situation?

It is impossible in the same sense as the equality 0 * y = 10 is impossible. The right-hand side, i.e., the value of multiplication, is not arbitrary. It is a function of the factors in the left-hand side, which means that it is fully determined when the factors are fixed. In the general case of a two-argument function, you need both arguments to determine its result, but multiplication has the property that if either argument is 0, the result cannot be anything other than 0. If you encounter 0 * y = 10, this is equivalent to a contradiction, from which you can derive everything. Similar to a truth table for implication, you won't find 0 * y = 10 for any y in the multiplication table.

You'll have to clarify what you mean by impossible. Otherwise, I think the use of "impossible" in your solution isn't relevant to my question.

What you described was an instance of a math statement A when A represents the true statement "0*y=0" for the compound statement A^(~A). Here, you posed a specific case of ~A, namely 0*y is not equal to 0 but is equal to some number other than 0 (in your case, equal to 10 -- let (~A) be statement B). A^(~A) is a contradiction and we can verify that with a truth table. 0*y=10 is a false statement, not an "impossible statement" (I don't mean to be a jerk but I have to be precise with terms here). 0*y=0 AND 0*y=10 is an example of a contradiction.

However, even though it is impossible to find y such that 0*y=10, the notion of impossibility you used there doesn't translate to the case I described. You're referring to a case of B in the context of the operator A^B. My question pertains to the compound statement created using the implication operator.


To recapitulate, I asked about the combination involving the implication operator. It's truth table is:
A B (A=>B)
T T T
T F F
F T T
F F T

The truth-table definition doesn't accommodate for when (A=>B) and A is false. I wonder why this is and what to do if a case like that ever arises. For example, what if I was given an If/then theorem (Say A=>B) that was actually false (that was thought to be true) and I were to apply a specific case of the theorem's antecedent (call this A) that I thought was true (but was actually false). According to the rule of inference (Modus Ponens), the consequent B MUST be true. However, Modus Ponens doesn't actually apply since A=>B that I was given was actually false. Is B still true or is it false?

This is puzzling to me because this situation is undefined according to the standard definition but can exist. A textbook gives a student a false math if/then theorem and the student makes a mistake and applies a false antecedent. He follows through with the inference and concludes the consequent is true. Is it really true?