T5.6 3 of 6
So, one might well proceed as follows.
Premise | 1. | [A=(B=L)]&[C=(X&(B>L))] |
Premise | 2. | A |
1 &E | 3. | A=(B=L) |
2,3 =E | 4. | B=L |
4 EQ | 5. | (B>L)&(L>B) |
5 &E | 6. | B>L |
So, we are done pretty easily.
There are a few rules of strategy that work well in most cases.
Rules of Strategy for SD Derivations:
Let's see this 1, 2, 3, 4 strategy at work on this one.
Suppose that we are asked to prove that the following argument is valid.
F>G
~GvH
F>H
We have been showing arguments valid by derivation throughout this chapter. Still let's get clear on the definition.
To show an argument valid in SD, simply derive its conclusion from it's premises.
Keep in mind that "derive" here means that the conclusion needs to be in the main derivation column and not in a subderivation. The subderivation is not nearly good enough: everything in a subderivation is suspect and there to be shown false!
OK, let's go through our reasoning.
Here's that argument again:
F>G
~GvH
F>H
And here's our strategy list applied to this one:
So, there is an easy way to do this one, but...
...let's suppose that we don't notice that HS and IM will work in two steps to finish by line 4! (You wouldn't see this easy way at first, would you? That takes practice. More on the easy way in a moment.)
So let's say we get to the "when all else fails" step. Well, then the derivation will get a bit longer, but you can just do some of the things we got used to in 5.1-5.4.
So, let's say we don't know what else to do to derive our goal 'F>H'.
In this case, you , scratch your head a few times, and seeing nothing else to do, go to 4: RD and the assumption of the opposite of your goal.
OK, let's see where we'd go from here.
Now, notice that the assumption on line 3 is like so many others we've seen. It's a negation of a conditional. Complex negations almost always are broken down by DM and that requires we first do IM.
But we made the assumption of '~(F>H)' on line 3 to show that it is false! We need a contradiction. So let's use 1 and 2 along with 6 to show '~(F>H)' could not be true.
But this is familiar stuff:
We just apply the standard rules to lines 6, then 7, and then 8.
It takes very little extra work from here to get the contradiction: 'H' and '~H' on lines 10 and 11.
(Notice that 'G' and '~~G' not don't contradict each other, a contradiction is always one sentence and its negation.
This looks complicated. But remember that this is made up of 1) the assumption for contradiction at line 3, 2) our usual way of handing a negation on lines 4-6, and 3) a few of our oldest rules to get the contradiction of 10 and 11.
Simple Way: But there is an easier way if we think about rules that lead to the goal and pattern matching for the two premises.
Let's see it!
This is simple, if you can see it.
So, what can we do at line 4 to finish this thing?