Proof: Equation a

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	(a + c) + d = (c + a) + d	(a + c) + d = (c + a) + d (sum_equation_pre_1)
2	((a + c) + d) + b = ((c + a) + d) + b	if (a + c) + d = (c + a) + d, then ((a + c) + d) + b = ((c + a) + d) + b (Additive Property of Equality)
3	((c + a) + d) + b = (c + a) + (d + b)	((c + a) + d) + b = (c + a) + (d + b) (Associative Property of Addition)
4	d + b = b + d	d + b = b + d (Commutative Property of Addition)
5	(c + a) + (d + b) = (c + a) + (b + d)	if d + b = b + d, then (c + a) + (d + b) = (c + a) + (b + d) (Substitution)
6	(c + a) + (b + d) = ((c + a) + b) + d	(c + a) + (b + d) = ((c + a) + b) + d (associative)
7	(c + a) + b = (b + c) + a	(c + a) + b = (b + c) + a (rearrange_sum_equal_3)
8	((c + a) + b) + d = ((b + c) + a) + d	if (c + a) + b = (b + c) + a, then ((c + a) + b) + d = ((b + c) + a) + d (Additive Property of Equality)
9	((a + c) + d) + b = ((b + c) + a) + d	if ((c + a) + b) + d = ((b + c) + a) + d and (c + a) + (b + d) = ((c + a) + b) + d and (c + a) + (d + b) = (c + a) + (b + d) and ((c + a) + d) + b = (c + a) + (d + b) and ((a + c) + d) + b = ((c + a) + d) + b, then ((a + c) + d) + b = ((b + c) + a) + d (Transitive Property Application 5)

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