715: Proposition 1 or Proposition 2 iff if Not Proposition 2, Proposition 1

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description/proof of that proposition 1 or proposition 2 iff if not proposition 2, proposition 1

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Topics

About: logic

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof
• 4: Note

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Starting Context

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Target Context

• The reader will have a description and a proof of the proposition that any proposition 1 or any proposition 2 if and only if if not the proposition 2, the proposition 1.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$p_1$: $\in \{\text{ the propositions }\}$
$p_2$: $\in \{\text{ the propositions }\}$
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Statements:
$p_1\vee p_2$
$⟺$
$\mathrm{\neg }{p}_2⟹p_1$
//

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2: Natural Language Description

For any propositions, $p_1$ and $p_2$, $p_1\vee p_2$ if and only if $\mathrm{\neg }{p}_2⟹p_1$.

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3: Proof

Whole Strategy: Step 1: suppose that $p_1\vee p_2$ is true and prove that $\mathrm{\neg }{p}_2⟹p_1$ is true; Step 2: suppose that $\mathrm{\neg }{p}_2⟹p_1$ is true and prove that $p_1\vee p_2$ is true.

Step 1:

Let us suppose that $p_1\vee p_2$ is true.

If $\mathrm{\neg }{p}_2$ is true, $p_2$ will be false, so, $p_1$ will be true (because $p_1\vee p_2$ is true), and so, $\mathrm{\neg }{p}_2⟹p_1$ is true.

Step 2:

Let us suppose that $\mathrm{\neg }{p}_2⟹p_1$ is true.

If $p_2$ is true, $p_1\vee p_2$ is true. If $p_2$ is false, $\mathrm{\neg }{p}_2$ is true, and $p_1$ is true (because $\mathrm{\neg }{p}_2⟹p_1$ is true), and so, $p_1\vee p_2$ is true. So, $p_1\vee p_2$ is true anyway.

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4: Note

Of course, also $\mathrm{\neg }{p}_1⟹p_2$ is equivalent, because $p_1\vee p_2⟺p_2\vee p_1$ and also because $\mathrm{\neg }{p}_1⟹p_2$ is the contraposition of $\mathrm{\neg }{p}_2⟹p_1$.

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References

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