Commutative Property Variation 1
Substitution 2
Substitute 6

Proof: Substitute 6

Let's prove the following theorem:

if the following are true:
  • a = ((b + c) + d) + e
  • d + e = f

then a = (b + c) + f

Proof:

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Given
1 a = ((b + c) + d) + e
2 d + e = f
Proof Table
# 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)
3 a = (b + c) + f if a = (b + c) + (d + e) and d + e = f, then a = (b + c) + f
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