Proof: Add 6 Numbers
Let's prove the following theorem:
((a + b) + c) + ((d + e) + f) = ((((a + b) + c) + d) + e) + f
Proof:
# | 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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