2

If I have something like the following can I use the add inference rule to add ~A. Does that cause a contradiction, or am I fine since it's if A and not A being directly declared?

1. (A ⊃ B) ⊃ C 
2. ~D ⊃ B             

3.suppose ~D
4. B            2,3 MP
5. B v ~A       2 Add
6. ~A v B       5 Comm
7. A ⊃ B        6 CE
8. C            1,7 MP
Close supposition
9. ~D ⊃ C
2

The inference from:

  1. B

to

  1. B v ~A

is valid.

We can prove this using a truth table,

For this truth table, we already know that B is true, so we just need to know that B v ~A is true in every instance to be able to add it.

B | A | ~A  | A v B 
T | T |  F  |   T
T | F |  T  |   T

As you can see, we can add anything with v in this way -- both A and not A work. (The same is not true of other operators).

The only error I see is that you're referencing the wrong premise. You should reference line 4 (4. B) rather than 2 (2. ~D -> B)

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