Cole A.
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Homework Statement
Prove that if x^2 + y = 13 and y \neq 4, then x \neq 3.
Homework Equations
N.A.
The Attempt at a Solution
The proof itself is simple enough: suppose x^2 + y = 13 and y \neq 4. Suppose for the sake of contradiction that x = 3. Then
<br /> \begin{align*}<br /> (3)^2 + y &= 13 \\<br /> y &= 4.<br /> \end{align*}
But this contradicts the knowledge that y \neq 4. Therefore, if x^2 + y = 13 and y \neq 4, then x \neq 3.
The problem I am having is understanding why this is logically valid. Would it be correct to say that, for the statements
<br /> \begin{align*}<br /> A &: x^2 + y = 13 \\<br /> B &: y = 4 \\<br /> C &: x = 3,<br /> \end{align*}<br />
what has been proven is below?
<br /> \begin{align*}<br /> &(A \wedge \neg B \wedge C) \rightarrow B \\<br /> \text{which is equivalent to}~ &(A \wedge \neg B) \rightarrow (C \rightarrow B) \\<br /> \text{which is equivalent to}~ &(A \wedge \neg B) \rightarrow (\neg B \rightarrow \neg C) \\<br /> \text{which is equivalent to}~ &(A \wedge \neg B \wedge \neg B) \rightarrow \neg C \\<br /> \text{which is equivalent to}~ &(A \wedge \neg B) \rightarrow \neg C.<br /> \end{align*}<br />
Is this the proper way to think about the validity of proof by contradiction? (Sorry if this is a dumb question, I'm not a mathematician. What I am finding hard to stomach is identifying x = 3 as the contradictory statement when there are actually three statements that were assumed to be true (and thus possible culprits of the contradiction)).