Proof: Add 6 Numbers

Let's prove the following theorem:

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

Proof:

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

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