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."