This problem feels like it should be simpler than I've made it so my ultimate question will be: is there a simpler way to do this? In logic we know that

```
A v B = B v A

```

But in natural deduction we use our v-Introductions, RAA, etc. to prove these equivalences. In the process of solving a practice problem, I encountered the need to prove this commutative property but am finding it surprisingly difficult. It seems to me that the proof will start out like this:

```
1. A v B            given
2.     ¬(B v A)     assume
3.     ¬B ^ ¬A      2, deMorgan's
4.     ¬A           3, ^-elimination
5.     ¬B           3, ^-elimination
6.     ¬A ^ ¬B      4, 5, ^-I
7.     ¬(A v B)     6, deMorgan's
?. B v A            2, 7 RAA

```

And now we find ourselves in a position where we must prove deMorgan's. Is there no simpler proof for A v B = B v A?