Transitive Property of Equality Variation 2
Multiplication Property 2

Proof: Inverse Product Theorem

Let's prove the following theorem:

if not (a = 0), then (1 / a) ⋅ a = 1

Proof:

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Given
1 not (a = 0)
Proof Table
# Claim Reason
1 a / a = 1 if not (a = 0), then a / a = 1
2 a / a = a ⋅ (1 / a) a / a = a ⋅ (1 / a)
3 a ⋅ (1 / a) = (1 / a) ⋅ a a ⋅ (1 / a) = (1 / a) ⋅ a
4 a / a = (1 / a) ⋅ a if a ⋅ (1 / a) = (1 / a) ⋅ a and a / a = a ⋅ (1 / a), then a / a = (1 / a) ⋅ a
5 (1 / a) ⋅ a = 1 if a / a = 1 and a / a = (1 / a) ⋅ a, then (1 / a) ⋅ a = 1
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