# Proof

(related to Lemma: Implication as a Disjunction)

### Context

• Let $x,y$ be two propositions.
• We want to show that $(x\Rightarrow y)\Longleftrightarrow (\neg x \vee y).$

### Hypothesis

• We are given the implication $x\Rightarrow y$.

### Implications

$$\models(x)$$ $$\models(y)$$ $$\models(x \Rightarrow y)$$
$$1$$ $$1$$ $$1$$
$$0$$ $$1$$ $$1$$
$$1$$ $$0$$ $$0$$
$$0$$ $$0$$ $$1$$
• Based on the truth tables of the negation and disjunction, the truth table of $(\neg x\vee y)$ is given by

$$\models(x)$$| $$\models(y)$$| $$\models(\neg x)$$| $$\models(\neg x \vee y)$$ $$1$$| $$1$$| $$0$$| $$1$$ $$0$$| $$1$$| $$1$$| $$1$$ $$1$$| $$0$$| $$0$$| $$0$$ $$0$$| $$0$$| $$1$$| $$1$$

### Conclusion

• Since the outcomes (columns to the right) of both truth tables are equal, we have shown the equivalence $(x\Rightarrow y)\Longleftrightarrow (\neg x \vee y).$

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### References

#### Bibliography

1. Mendelson Elliott: "Theory and Problems of Boolean Algebra and Switching Circuits", McGraw-Hill Book Company, 1982