Formal Logic 101 – Part 11: Replacement Rules (Part 2)
Welcome back. Today we’ll be picking up where we left off with more replacement rules.
Our symbolism would be as follows:
1. (L^S)→ F Premise (P)
2. S^L P
Now, instinctively, I think we understand that “L^S” and “S^L” mean the same thing, but as was stated in the introduction, formal logic requires you to show your work. Every step must be documented. It’s not enough to assume that those two things are equivalent, we must show why our assertion is justified.
Or we can use Commutation as a shortcut as follows:
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. L^S 2 comm.
4. F 1,3 M.P.
You may recall that in the last post, I mentioned that replacement rules can be used on parts of lines. That means it would be equally correct to say:
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. (S^L)→F 1 comm.
4. F 2,3 M.P.
Unlike inference rules where you need to get the operators out of the parentheses before you can do anything with them, you can use replacement rules even when the operator is still inside parentheses (this is what I mean by saying it can be used on "parts of lines"). Remember, this is because we’re not using replacement rules to reach any new conclusions, we’re just rephrasing what we already have.
1. A→B P – Prove ~B→~A
2. ~B Assumption
3. ~A 1,2 M.T.
4. ~B→~A 2-3 C.P.
So contraposition lets us shave out two steps by simply going:
1. A→B Premise (P) – (prove “~B→~A”)
2. ~B→~A 1 Contrap.
That puts us half-way through our replacement rules. Five down, five to go. I will once again not include practice problems for this one, as we will be covering the most commonly used replacement rules next time, so I think it best to hold off until we have them in our toolbox as well. So we’ll have practice arguments again next time.
Commutation (Comm.)
Commutation is a handy step for dealing with “And” or “Or” statements, where the statement you have is close to what you would need to complete another step (such Modus Ponens, for example) but is not quite exactly right. I think the easiest way to explain this would be by example.Consider the premises:
If the Sun is very Large but looks very Small, then it must be Far away.
The Sun looks very Small, but it is very Large.
If the Sun is very Large but looks very Small, then it must be Far away.
The Sun looks very Small, but it is very Large.
Our symbolism would be as follows:
1. (L^S)→ F Premise (P)
2. S^L P
Now, instinctively, I think we understand that “L^S” and “S^L” mean the same thing, but as was stated in the introduction, formal logic requires you to show your work. Every step must be documented. It’s not enough to assume that those two things are equivalent, we must show why our assertion is justified.
We can do that with the rules we have so far as such:
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. S 2 simp.
4. L 2 simp.
5. L^S 3,4 conj.
6. F 1,5 M.P.
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. S 2 simp.
4. L 2 simp.
5. L^S 3,4 conj.
6. F 1,5 M.P.
Or we can use Commutation as a shortcut as follows:
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. L^S 2 comm.
4. F 1,3 M.P.
You may recall that in the last post, I mentioned that replacement rules can be used on parts of lines. That means it would be equally correct to say:
1. (L^S)→ F P
2. S^L P – (prove “F”)
3. (S^L)→F 1 comm.
4. F 2,3 M.P.
Unlike inference rules where you need to get the operators out of the parentheses before you can do anything with them, you can use replacement rules even when the operator is still inside parentheses (this is what I mean by saying it can be used on "parts of lines"). Remember, this is because we’re not using replacement rules to reach any new conclusions, we’re just rephrasing what we already have.
Contraposition (Contrap.)
I tend to think of Contraposition as kind of "half-way" between Modus Tollens and a Hypothetical Syllogism. Contraposition states that “A→B” means the same thing as “~B→~A”. This can be proven with a simple conditional proof, as follows:1. A→B P – Prove ~B→~A
2. ~B Assumption
3. ~A 1,2 M.T.
4. ~B→~A 2-3 C.P.
So contraposition lets us shave out two steps by simply going:
1. A→B Premise (P) – (prove “~B→~A”)
2. ~B→~A 1 Contrap.
That puts us half-way through our replacement rules. Five down, five to go. I will once again not include practice problems for this one, as we will be covering the most commonly used replacement rules next time, so I think it best to hold off until we have them in our toolbox as well. So we’ll have practice arguments again next time.
Comments
Post a Comment