MandysNotes

Sunday, 06 April 2014 00:00

Truth Values for Implication

By  Gideon
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Suppose your friend tells you:

"If the animal I am thinking of is duck-billed, then it is also a mammal."

Assume that he is telling you the truth (for now). 

He is telling you that the statement:

\[ A \to B \]

is true. 

Logically, this is the same as saying that the statement:

\[ ( \lnot A) \cup B\]

is true. 

That is, the animal must be somewhere in the green area in the figure below, and cannot be in the red area. 

NotAorBtrueGR

 Now supose he tells you that the animal is duck-billed; that is, he tells you that \(A\) is true. 

Therefore the animal he is thinking of must be in the green part of the figure below, and cannot be in the red part. 

AtrueGR1

 Putting these results together we see that we must have:

AandBtrueGR1

That is, the animal must be both duck-billed and a mammal (it must be a platypus).

 Now suppose that your friend tells you that the animal is not a mammal. 

That is, he tells you that \(B\) is false. 

That means that the animal must be in the green area in the figure below, and cannot be in the red area.

Bfalse1

 Combining this with the fact that \(A \to B\) is true:

NotAorBtrueGR

We see that both \(A\) and \(B\) must be false.

AfalseBfalseGR1

 

What if your friend tells you that \(A \to B\) is true, but that \(A\) is false?

That is, that if the animal is duck-billed then it is a mammal, and that the animal is not duck-billed.

There are two possibilities in this case:

First, the animal could be a mammal.

NotAGR1 

Second, it is possible to have the situation already considered, such that \(A \to B\) is true, but both \(A\) and \(B\) are false

AfalseBfalseGR1

Thus we see that knowing that \(A \to B\) is true, and that \(A\) is false, is not enough to decide whether or not \(B\) is true. 

 

Finally, suppose that your friend is lying about \( A \to B\) being true;

that is, assume that \(A \to B = ( \lnot A) \cup B \) is false. 

That means that the animal must be in the green area in the figure below, and cannot be in the red area.

NotAorBfalseGR1

 We see that in this case \(A\) is true and \(B\) is false. 

Putting all these results together we obtain the truth table for implication:

NotAorBTT2

Note that \(A \to B\) is true unless \(A\) is true and \(B\) is false. 

 

Read 1483 times Last modified on Sunday, 06 April 2014 23:44
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