Metamath Proof Explorer


Theorem resum2sqorgt0

Description: The sum of the square of two real numbers is greater than zero if at least one of the real numbers is nonzero. (Contributed by AV, 26-Feb-2023)

Ref Expression
Hypothesis resum2sqcl.q ⊢ Q = A 2 + B 2
Assertion resum2sqorgt0 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 → 0 < Q

Proof

Step Hyp Ref Expression
1 resum2sqcl.q ⊢ Q = A 2 + B 2
2 1 resum2sqgt0 ⊢ A ∈ ℝ ∧ A ≠ 0 ∧ B ∈ ℝ → 0 < Q
3 2 ex ⊢ A ∈ ℝ ∧ A ≠ 0 → B ∈ ℝ → 0 < Q
4 3 expcom ⊢ A ≠ 0 → A ∈ ℝ → B ∈ ℝ → 0 < Q
5 4 com23 ⊢ A ≠ 0 → B ∈ ℝ → A ∈ ℝ → 0 < Q
6 eqid ⊢ B 2 + A 2 = B 2 + A 2
7 6 resum2sqgt0 ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → 0 < B 2 + A 2
8 1 breq2i ⊢ 0 < Q ↔ 0 < A 2 + B 2
9 resqcl ⊢ A ∈ ℝ → A 2 ∈ ℝ
10 9 adantl ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → A 2 ∈ ℝ
11 10 recnd ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → A 2 ∈ ℂ
12 resqcl ⊢ B ∈ ℝ → B 2 ∈ ℝ
13 12 ad2antrr ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → B 2 ∈ ℝ
14 13 recnd ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → B 2 ∈ ℂ
15 11 14 addcomd ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → A 2 + B 2 = B 2 + A 2
16 15 breq2d ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → 0 < A 2 + B 2 ↔ 0 < B 2 + A 2
17 8 16 bitrid ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → 0 < Q ↔ 0 < B 2 + A 2
18 7 17 mpbird ⊢ B ∈ ℝ ∧ B ≠ 0 ∧ A ∈ ℝ → 0 < Q
19 18 ex ⊢ B ∈ ℝ ∧ B ≠ 0 → A ∈ ℝ → 0 < Q
20 19 expcom ⊢ B ≠ 0 → B ∈ ℝ → A ∈ ℝ → 0 < Q
21 5 20 jaoi ⊢ A ≠ 0 ∨ B ≠ 0 → B ∈ ℝ → A ∈ ℝ → 0 < Q
22 21 3imp31 ⊢ A ∈ ℝ ∧ B ∈ ℝ ∧ A ≠ 0 ∨ B ≠ 0 → 0 < Q