Algebra 1 / Chapter 2: Multiplication / Reordering Theorems

Proof: Move Terms 2 and 4 Theorem

Let's prove the following theorem:

((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c

This example shows that we can reorder terms in any way we want.

In step 2 - 4, we show that:

(a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c)

And in step 6, we use the Swap Inner Terms theorem to claim that:

((b ⋅ a) ⋅ d) ⋅ c = ((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 = b ⋅ a a ⋅ b = b ⋅ a
4 (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) if a ⋅ b = b ⋅ a and c ⋅ d = d ⋅ c, then (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c)
5 (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c
6 ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c
7 ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c if ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) and (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) and (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c and ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c, then ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c
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