Proof: Move Terms 2 and 4 Theorem

Let's prove the following theorem:

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:

Proof Table

#	Claim	Reason
1	((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d)	((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d) (Associative Property of Multiplication)
2	c ⋅ d = d ⋅ c	c ⋅ d = d ⋅ c (Commutative Property of Multiplication)
3	a ⋅ b = b ⋅ a	a ⋅ b = b ⋅ a (Commutative Property of Multiplication)
4	(a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c)	if c ⋅ d = d ⋅ c and a ⋅ b = b ⋅ a, then (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) (Substitute Two Variables Theorem)
5	(b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c	(b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c (associative_property_of_multiplication_2)
6	((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c	((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c (reorder_terms_7)
7	((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c	if ((b ⋅ a) ⋅ d) ⋅ c = ((b ⋅ d) ⋅ a) ⋅ c and (b ⋅ a) ⋅ (d ⋅ c) = ((b ⋅ a) ⋅ d) ⋅ c and (a ⋅ b) ⋅ (c ⋅ d) = (b ⋅ a) ⋅ (d ⋅ c) and ((a ⋅ b) ⋅ c) ⋅ d = (a ⋅ b) ⋅ (c ⋅ d), then ((a ⋅ b) ⋅ c) ⋅ d = ((b ⋅ d) ⋅ a) ⋅ c (Transitive Property Application 4)

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