Proof: Substitute First Term

Let's prove the following theorem:

if the following are true:

(a + b) + c = d
a = e

then (e + b) + c = d

Proof:

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Given

1	(a + b) + c = d
2	a = e

Proof Table

#	Claim	Reason
1	(a + b) + c = a + (b + c)	(a + b) + c = a + (b + c) (Associative Property of Addition)
2	(b + c) + a = a + (b + c)	(b + c) + a = a + (b + c) (Commutative Property of Addition)
3	a + (b + c) = (a + b) + c	if (a + b) + c = a + (b + c), then a + (b + c) = (a + b) + c (Symmetric Property of Equality)
4	(b + c) + a = (a + b) + c	if (b + c) + a = a + (b + c) and a + (b + c) = (a + b) + c, then (b + c) + a = (a + b) + c (Transitive Property of Equality)
5	(b + c) + a = d	if (b + c) + a = (a + b) + c and (a + b) + c = d, then (b + c) + a = d (Transitive Property of Equality)
6	(b + c) + e = d	if (b + c) + a = d and a = e, then (b + c) + e = d (Substitution Example 10)
7	e + (b + c) = d	if (b + c) + e = d, then e + (b + c) = d (Commutative Property Example 2)
8	(e + b) + c = e + (b + c)	(e + b) + c = e + (b + c) (Associative Property of Addition)
9	(e + b) + c = d	if (e + b) + c = e + (b + c) and e + (b + c) = d, then (e + b) + c = d (Transitive Property of Equality)

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