- #1

- 479

- 12

Hi all,

There's this proof that I've been trying to wrap my head around but it just doesn't seem to sink in. I've attached a screenshot below. Many thanks in advance!

Consider Case 1. There is a step that goes

$$\text{Then} \ |r| = r$$

$$Then -|r| \leq |r| \ \text{and} \ r \leq |r|$$

Why is this the case? This seems to imply that because ##|r|=r##, then ##r \leq |r|##. Is this because of the "generalisation" rule of inference that goes

$$p$$

$$\text{Therefore} \ p \vee q$$

Where ##p = |r| = r## and ##q = |r| > r##? If so, why not write ##q = |r| < r## and get a completely different result altogether?

There's this proof that I've been trying to wrap my head around but it just doesn't seem to sink in. I've attached a screenshot below. Many thanks in advance!

Consider Case 1. There is a step that goes

$$\text{Then} \ |r| = r$$

$$Then -|r| \leq |r| \ \text{and} \ r \leq |r|$$

Why is this the case? This seems to imply that because ##|r|=r##, then ##r \leq |r|##. Is this because of the "generalisation" rule of inference that goes

$$p$$

$$\text{Therefore} \ p \vee q$$

Where ##p = |r| = r## and ##q = |r| > r##? If so, why not write ##q = |r| < r## and get a completely different result altogether?