Proof: Square

Let's prove the following theorem:

((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = (aa) + (bb)

Proof:

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Proof Table
# Claim Reason
1 ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb)
2 ((ab) ⋅ 2) + ((ab) ⋅ (-2)) = 0 ((ab) ⋅ 2) + ((ab) ⋅ (-2)) = 0
3 ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (((ab) ⋅ 2) + ((ab) ⋅ (-2))) + ((aa) + (bb)) ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (((ab) ⋅ 2) + ((ab) ⋅ (-2))) + ((aa) + (bb))
4 ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = 0 + ((aa) + (bb)) if ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (((ab) ⋅ 2) + ((ab) ⋅ (-2))) + ((aa) + (bb)) and ((ab) ⋅ 2) + ((ab) ⋅ (-2)) = 0, then ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = 0 + ((aa) + (bb))
5 0 + ((aa) + (bb)) = (aa) + (bb) 0 + ((aa) + (bb)) = (aa) + (bb)
6 ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (aa) + (bb) if ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = 0 + ((aa) + (bb)) and 0 + ((aa) + (bb)) = (aa) + (bb), then ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (aa) + (bb)
7 ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb))
8 ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) if ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)), then ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb)
9 ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = (aa) + (bb) if ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) and ((((ab) ⋅ 2) + ((ab) ⋅ (-2))) + (aa)) + (bb) = (aa) + (bb), then ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = (aa) + (bb)
10 ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = (aa) + (bb) if ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) and ((((ab) ⋅ 2) + (aa)) + ((ab) ⋅ (-2))) + (bb) = (aa) + (bb), then ((ab) ⋅ 2) + (((aa) + ((ab) ⋅ (-2))) + (bb)) = (aa) + (bb)
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