Proof: Algebra Square Sum

Let's prove the following theorem:

if the following are true:
  • x = (a ⋅ a) + (b ⋅ b)
  • m = a
  • n = b

then x = (n ⋅ n) + (m ⋅ m)

Proof:

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

Given
1 x = (a ⋅ a) + (b ⋅ b)
2 m = a
3 n = b
Proof Table
# Claim Reason
1 m ⋅ m = a ⋅ a if m = a, then m ⋅ m = a ⋅ a
2 x = (m ⋅ m) + (b ⋅ b) if m ⋅ m = a ⋅ a and x = (a ⋅ a) + (b ⋅ b), then x = (m ⋅ m) + (b ⋅ b)
3 n ⋅ n = b ⋅ b if n = b, then n ⋅ n = b ⋅ b
4 x = (m ⋅ m) + (n ⋅ n) if n ⋅ n = b ⋅ b and x = (m ⋅ m) + (b ⋅ b), then x = (m ⋅ m) + (n ⋅ n)
5 (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m) (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m)
6 x = (n ⋅ n) + (m ⋅ m) if (m ⋅ m) + (n ⋅ n) = (n ⋅ n) + (m ⋅ m) and x = (m ⋅ m) + (n ⋅ n), then x = (n ⋅ n) + (m ⋅ m)

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