Metamath Proof Explorer


Theorem 2sqreulem2

Description: Lemma 2 for 2sqreu etc. (Contributed by AV, 25-Jun-2023)

Ref Expression
Assertion 2sqreulem2 ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → A 2 + B 2 = A 2 + C 2 → B = C

Proof

Step Hyp Ref Expression
1 nn0cn ⊢ A ∈ ℕ 0 → A ∈ ℂ
2 1 sqcld ⊢ A ∈ ℕ 0 → A 2 ∈ ℂ
3 2 3ad2ant1 ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → A 2 ∈ ℂ
4 nn0cn ⊢ B ∈ ℕ 0 → B ∈ ℂ
5 4 sqcld ⊢ B ∈ ℕ 0 → B 2 ∈ ℂ
6 5 3ad2ant2 ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → B 2 ∈ ℂ
7 nn0cn ⊢ C ∈ ℕ 0 → C ∈ ℂ
8 7 sqcld ⊢ C ∈ ℕ 0 → C 2 ∈ ℂ
9 8 3ad2ant3 ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → C 2 ∈ ℂ
10 3 6 9 addcand ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → A 2 + B 2 = A 2 + C 2 ↔ B 2 = C 2
11 nn0sq11 ⊢ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → B 2 = C 2 ↔ B = C
12 11 biimpd ⊢ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → B 2 = C 2 → B = C
13 12 3adant1 ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → B 2 = C 2 → B = C
14 10 13 sylbid ⊢ A ∈ ℕ 0 ∧ B ∈ ℕ 0 ∧ C ∈ ℕ 0 → A 2 + B 2 = A 2 + C 2 → B = C