Proof: Reorder Terms

Let's prove the following theorem:

((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c

Proof:

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Proof Table
# Claim Reason
1 ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d)
2 c ⋅ d = d ⋅ c c ⋅ d = d ⋅ c
3 (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c) if c ⋅ d = d ⋅ c, then (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c)
4 a ⋅ b = b ⋅ a a ⋅ b = b ⋅ a
5 (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) if (a ⋅ b) ⋅ (c ⋅ d) = (a ⋅ b) ⋅ (d ⋅ c) and a ⋅ b = b ⋅ a, then (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c)
6 (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c
7 (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a
8 ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c if (b ⋅ a) ⋅ d = (b ⋅ d) ⋅ a, then ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c
9 (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c if (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c and ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c, then (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c
10 (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c if (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) and (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ d) ⋅ a) ⋅ c, then (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c
11 ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c if ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) and (a ⋅ b) ⋅ (c ⋅ d) = ((b ⋅ d) ⋅ a) ⋅ c, then ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c

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