Proof: Manipulation 3

Let's prove the following theorem:

(a ⋅ (-1)) ⋅ (a ⋅ (-1)) = a ⋅ a

Proof:

View as a tree | View dependent proofs | Try proving it

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

Comments

Please log in to add comments