T5.4 p2 Negation Rules

T5.4 2 of 4

Again, our newest rule is:

input:

output:

	P Q ~Q

~P

The final rule of SD, negation elimination, is very similar to negation introduction. But for negation elimination the output will be P while the input is ~P:

input:

output:

	~P Q ~Q

P

The idea here is that to prove some statement P we may assume the negation of P, show that the negation cannot be true because it leads to contradiction, so conclude that P is true.

Let's look at a derivation making use of a negation rule...

From the premise '(~A=B)&~B' deduce 'A'. We will say that 'A' is our goal.

Helpful but Optional: Again we write down the goal sentence toward the end of the derivation with question marks -- thus we keep in mind that it is merely a goal and not yet justified.

It is sometimes difficult to know how to start a derivation. Hint: think about ~E and so assume the negation of the goal sentence. (Note: there are lots of other equally good ways to begin.)

Now, we are working within a subderivation. We assumed '~A' and hopefully will be able to derive some contradiction. (Why?) Toward this intermediate goal we now apply &E to line 1 for the first time...

Now we may use =E.

Finally, we can get our contradiction with another application of &E...can you guess how?

The assumption of line 2, '~A', has led to contradiction. So, 'A' must be true. Thus we have terminated the subderivation and justified 'A'.

Replay the demonstration.
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