Proof: Add 6 Numbers

Let's prove the following theorem:

Proof:

Proof Table

#	Claim	Reason
1	((a + b) + c) + ((d + e) + f) = (((a + b) + c) + (d + e)) + f	((a + b) + c) + ((d + e) + f) = (((a + b) + c) + (d + e)) + f (associative)
2	((a + b) + c) + (d + e) = (((a + b) + c) + d) + e	((a + b) + c) + (d + e) = (((a + b) + c) + d) + e (associative)
3	(((a + b) + c) + (d + e)) + f = ((((a + b) + c) + d) + e) + f	if ((a + b) + c) + (d + e) = (((a + b) + c) + d) + e, then (((a + b) + c) + (d + e)) + f = ((((a + b) + c) + d) + e) + f (Additive Property of Equality)
4	((a + b) + c) + ((d + e) + f) = ((((a + b) + c) + d) + e) + f	if (((a + b) + c) + (d + e)) + f = ((((a + b) + c) + d) + e) + f and ((a + b) + c) + ((d + e) + f) = (((a + b) + c) + (d + e)) + f, then ((a + b) + c) + ((d + e) + f) = ((((a + b) + c) + d) + e) + f (Transitive Property of Equality)

Comments

Please log in to add comments