Proof: Subtract Both Sides 2

Let's prove the following theorem:

if a = b + c, then a + (b ⋅ (-1)) = c

Proof:

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Given
1 a = b + c
Proof Table
# Claim Reason
1 b + c = c + b b + c = c + b
2 a = c + b if a = b + c and b + c = c + b, then a = c + b
3 a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1)) if a = c + b, then a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1))
4 (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1)))
5 a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) if a + (b ⋅ (-1)) = (c + b) + (b ⋅ (-1)) and (c + b) + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))), then a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1)))
6 b + (b ⋅ (-1)) = 0 b + (b ⋅ (-1)) = 0
7 c + (b + (b ⋅ (-1))) = c + 0 if b + (b ⋅ (-1)) = 0, then c + (b + (b ⋅ (-1))) = c + 0
8 a + (b ⋅ (-1)) = c + 0 if a + (b ⋅ (-1)) = c + (b + (b ⋅ (-1))) and c + (b + (b ⋅ (-1))) = c + 0, then a + (b ⋅ (-1)) = c + 0
9 c + 0 = c c + 0 = c
10 a + (b ⋅ (-1)) = c if a + (b ⋅ (-1)) = c + 0 and c + 0 = c, then a + (b ⋅ (-1)) = c

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