Does Proving a Contradiction Imply a Logical Relationship?

[Horse]

Homework Statement

Prove or disprove:

"If you can prove $$( y \wedge \neg c ) \rightarrow Contradiction$$, then
$$y \rightarrow c$$ must be right."

Homework Equations

My teacher used the sign $$\wedge$$, instead of $$\vee$$, like:

"If $$( a \wedge b \wedge \neg c ) \rightarrow Contradiction$$, then $$a \wedge b \rightarrow c$$ must be right."

I feel it is not right.

The Attempt at a Solution

I proved in my replies:

"If you can prove $$( a \wedge b \vee \neg c ) \rightarrow Contradiction$$, then
$$a \wedge b \rightarrow c$$ must be right."

I used the facts in my proof:

$$a \wedge \neg b = \neg ( a \rightarrow b ) = ( a \not \rightarrow b ) \not = \neg b \rightarrow \neg a \not = \neg a \not \rightarrow \neg b$$

 

[hatsoff]

I'm pretty sure you're either missing something or else you have mistyped it. (*) does not follow from the two contradictions given.

 

[Horse]

I'm pretty sure you're either missing something or else you have mistyped it. (*) does not follow from the two contradictions given.

Let's simplify. I know for sure:$$\neg ( a \rightarrow b ) = a \wedge \neg b$$

So

$$( a \rightarrow b ) = \neg a \vee b$$

If I want to prove $$( a \rightarrow b )$$, then the finding $$a \wedge \neg b \rightarrow Contradiction$$ will prove it. Please, notice that $$\neg ( a \rightarrow b ) \rightarrow Contradiction$$ because the two statements are equivalent. As $$\neg ( a \rightarrow b )$$ and $$( a \rightarrow b )$$ cannot be true at the same time, the conclusion must be valid.Let's compare its logic to the logic in the case:

The Argument:

$$a \wedge b \rightarrow c$$ (*)

Its contradictions:

$$\neg a \wedge b \wedge \neg c \rightarrow Contradiction$$

$$a \wedge \neg b \wedge \neg c \rightarrow Contradiction$$

We notice that the argument is:
$$a \wedge b \rightarrow c = \neg ( a \rightarrow \neg b ) \rightarrow c$$

So $$\neg ( \neg ( a \rightarrow \neg b ) \rightarrow c ) \rightarrow Contradiction$$ must prove it, by the logic above this reply.

Let's write its part differently:

$$\neg ( \neg ( a \rightarrow \neg b ) \rightarrow c ) = \neg ( a \rightarrow \neg b ) \not \rightarrow c = a \wedge b \not \rightarrow c$$

So I need to find that:

$$( a \wedge b \not \rightarrow c ) \rightarrow Contradiction = ( \neg a \vee \neg b \rightarrow \neg c ) \rightarrow Contradiction$$

Let's call $$d = \neg a \vee \neg b$$. So

$$( d \rightarrow \neg c ) \rightarrow Contradiction$$

Let's use again:

$$( a \rightarrow b ) = \neg a \vee b$$

So it becomes:
$$( d \rightarrow \neg c ) \rightarrow Contradiction = ( \neg d \vee \neg c ) \rightarrow Contradiction$$

It is equivalent to:

$$( a \wedge b \vee \neg c ) \rightarrow Contradiction$$

Conclusion

My teacher probably had something wrong. It should be right:

$$( a \wedge b \vee \neg c ) \rightarrow Contradiction$$

 

[Horse]

New Problem

Prove $$a \vee b \rightarrow c$$

Conjecture, according to my last proof:

"If you can prove $$( a \vee b \vee \neg c ) \rightarrow Contradiction$$, then
$$a \vee b \rightarrow c$$ must be right."

It is similar to the last proof by contradiction:

"If you can prove $$( a \wedge b \vee \neg c ) \rightarrow Contradiction$$, then
$$a \wedge b \rightarrow c$$ must be right."

 

[Horse]

Prove $$a \vee b \rightarrow c$$

Conjecture, according to my last proof:

"If you can prove $$( a \vee b \vee \neg c ) \rightarrow Contradiction$$, then
$$a \vee b \rightarrow c$$ must be right."

It is similar to the last proof by contradiction:

"If you can prove $$( a \wedge b \vee \neg c ) \rightarrow Contradiction$$, then
$$a \wedge b \rightarrow c$$ must be right."

Let's analyse them. Let $$p = a \vee b$$ and $$y = a \wedge b$$. So the problems become:

"If you can prove $$( p \vee \neg c ) \rightarrow Contradiction$$, then
$$p \rightarrow c$$ must be right."

"If you can prove $$( y \vee \neg c ) \rightarrow Contradiction$$, then
$$y \rightarrow c$$ must be right."

I think the problem is now solved, because you can see it is basically of the same form.

 

[Horse]

"If you can prove $$( y \vee \neg c ) \rightarrow Contradiction$$, then
$$y \rightarrow c$$ must be right."

Does the following method work?

"If you can prove $$( y \wedge \neg c ) \rightarrow Contradiction$$, then
$$y \rightarrow c$$ must be right."