- #1
guguma
- 51
- 5
The "Implies" connective
I have a big problem understanding the logic behind the implies connective.
[tex]P \Longrightarrow Q[/tex]
The truth table for this is the same as the truth table for Q OR NOT P.
The Problem is that I cannot wrap my mind around the fact that the implies statement is equivalent to Q OR NOT P. Writing truth tables for the AND OR NOT connectives is intuitive. But for the implies statement I think the truth value of the statement when P is False is accepted by convention.
What I understand from IMPLIES is that when I say
[tex]P \Longrightarrow Q[/tex]
It means Q follows from P, it means that the Truth of P makes Q also True, and When P is True and Q is False then My implies statement is False. That is OK up to this point.
Now when P is False I immediately assume [tex]P \Longrightarrow Q[/tex] is True! That is the problem. Shouldn't it be indeterminate?
And for the equivalence of Q OR NOT P, this statement looks at two unrelated statements Q and P I am looking for the truth of Q or the truth of NOT P for my statement is correct, but
this [tex]P \Longrightarrow Q[/tex] talks about two statements which are related. Truth of Q should follow from the truth of P.
Example:
[tex]S = T \Longrightarrow \left( S\cap T = S \cup T \right)[/tex]
Assume P is the left hand statement and Q is the right hand statement
Now if [tex]S=T[/tex] is true, this statement is true unless the right hand side is false. I understand that. But when [tex]S \neq T[/tex] this statement is still true.
When we are following the logic here we are using the definitions of the equality, intersection and the union of sets. From that I can only conclude the first two results assuming S = T It follows that Q is True so this statement is True. But if I do not assume S = T then Q is also False so this makes the statement still True due to truth table convention of the implies statement. But the second result does not show that Q followed from P. Take this:
[tex]S = T \Longrightarrow \left( S\cap T \neq S \cup T \right)[/tex]
When P is False Q is True, so this statement is True too. Now how come both statements in the two examples are true?
Please help me with this, I am sure that I am overlooking something and I feel very stupid because I cannot wrap my mind around this.
I have a big problem understanding the logic behind the implies connective.
[tex]P \Longrightarrow Q[/tex]
The truth table for this is the same as the truth table for Q OR NOT P.
The Problem is that I cannot wrap my mind around the fact that the implies statement is equivalent to Q OR NOT P. Writing truth tables for the AND OR NOT connectives is intuitive. But for the implies statement I think the truth value of the statement when P is False is accepted by convention.
What I understand from IMPLIES is that when I say
[tex]P \Longrightarrow Q[/tex]
It means Q follows from P, it means that the Truth of P makes Q also True, and When P is True and Q is False then My implies statement is False. That is OK up to this point.
Now when P is False I immediately assume [tex]P \Longrightarrow Q[/tex] is True! That is the problem. Shouldn't it be indeterminate?
And for the equivalence of Q OR NOT P, this statement looks at two unrelated statements Q and P I am looking for the truth of Q or the truth of NOT P for my statement is correct, but
this [tex]P \Longrightarrow Q[/tex] talks about two statements which are related. Truth of Q should follow from the truth of P.
Example:
[tex]S = T \Longrightarrow \left( S\cap T = S \cup T \right)[/tex]
Assume P is the left hand statement and Q is the right hand statement
Now if [tex]S=T[/tex] is true, this statement is true unless the right hand side is false. I understand that. But when [tex]S \neq T[/tex] this statement is still true.
When we are following the logic here we are using the definitions of the equality, intersection and the union of sets. From that I can only conclude the first two results assuming S = T It follows that Q is True so this statement is True. But if I do not assume S = T then Q is also False so this makes the statement still True due to truth table convention of the implies statement. But the second result does not show that Q followed from P. Take this:
[tex]S = T \Longrightarrow \left( S\cap T \neq S \cup T \right)[/tex]
When P is False Q is True, so this statement is True too. Now how come both statements in the two examples are true?
Please help me with this, I am sure that I am overlooking something and I feel very stupid because I cannot wrap my mind around this.