 #1
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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?
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