A few basic questions in logic from Zorich's mathematical analysis Let $A$ and $B$ be two mathematical statements, each of which can either be true, or false. The first problem is to show that:


  If $A$ is false, then $A \implies B$ is always true $\cdots\cdots(\star)$


Now I can consider $A:\{x&gt;0\}$ and $B:\{x&gt;5\}$. Now suppose we observe that $x=-2$. Then $A$ is false. But we never have $A \implies B$. Is the problem wrongly stated or am I missing something?

Two problems in the same line are to show that:

$(i)~~(A \implies B)\implies (not~~A)~~or~~B$

$(ii)~~not~~(A \implies B)\implies A~~and~~(not~~B)$

I can solve these two if I can show $(\star)$. I can sense that there is some vacuously true argument coming up but the counter example (?) is bugging me. Thanks in advance for any help.
Part of the problem is that you are parameterizing your propositions, which is a bit premature [this is only propositional calculus - later comes predicate calculus].  What I mean is: When you write 'A : {x > 0}' you are (I assume) asserting that A is the property 'x > 0'.  In other words, you are parameterizing A - what you really want to write is A(x).  But this is neither true nor false until you specify exactly what x is [similarly for B(x)].  So let us take a few examples:

x = -2 : Then A(-2) is the property '-2 > 0' and B(-2) is the property '-2 > 5'.  Neither of these is true.  So 'A(-2) ==> B(-2)' is vacuously true.

x = 2 : Then A(2) is '2 > 0' [true] and B(2) is '2 > 5' [false]  Hence 'A(2) ==> B(2)' is false.

x = 6 : Then A(6) is '6 > 0' [true] and B(6) is '6 > 5' [true]  Hence 'A(6) ==> B(6)' is true.

[I assume the cases like 'x=2' leads to your assertion 'we never have A==>B' but the cases like 'x=6' argue against that assertion]

To reiterate: When you try to define a property as a relation involving an unknown, you are not defining a proposition.  For the definition to be a proposition [i.e to have a definite truth-value], it must be completely defined.
Part of the problem is that you are parameterizing your propositions, which is a bit premature [this is only propositional calculus - later comes predicate calculus].  What I mean is: When you write 'A : {x > 0}' you are (I assume) asserting that A is the property 'x > 0'.  In other words, you are parameterizing A - what you really want to write is A(x).  But this is neither true nor false until you specify exactly what x is [similarly for B(x)].  So let us take a few examples:

x = -2 : Then A(-2) is the property '-2 > 0' and B(-2) is the property '-2 > 5'.  Neither of these is true.  So 'A(-2) ==> B(-2)' is vacuously true.

x = 2 : Then A(2) is '2 > 0' [true] and B(2) is '2 > 5' [false]  Hence 'A(2) ==> B(2)' is false.

x = 6 : Then A(6) is '6 > 0' [true] and B(6) is '6 > 5' [true]  Hence 'A(6) ==> B(6)' is true.

[I assume the cases like 'x=2' leads to your assertion 'we never have A==>B' but the cases like 'x=6' argue against that assertion]

To reiterate: When you try to define a property as a relation involving an unknown, you are not defining a proposition.  For the definition to be a proposition [i.e to have a definite truth-value], it must be completely defined.