Proof: Substitute 6 Pre
Let's prove the following theorem:
if a = ((b + c) + d) + e, then a = (b + c) + (d + e)
Proof:
Given
1 | a = ((b + c) + d) + e |
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# | Claim | Reason |
---|---|---|
1 | ((b + c) + d) + e = (b + c) + (d + e) | ((b + c) + d) + e = (b + c) + (d + e) |
2 | a = (b + c) + (d + e) | if a = ((b + c) + d) + e and ((b + c) + d) + e = (b + c) + (d + e), then a = (b + c) + (d + e) |
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