Metamath Proof Explorer


Theorem isdomn5

Description: The equivalence between the right conjuncts in the right hand sides of isdomn and isdomn2 , in predicate calculus form. (Contributed by SN, 16-Sep-2024)

Ref Expression
Assertion isdomn5 ⊢ ∀ a ∈ B ∀ b ∈ B a · ˙ b = 0 ˙ → a = 0 ˙ ∨ b = 0 ˙ ↔ ∀ a ∈ B ∖ 0 ˙ ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙

Proof

Step Hyp Ref Expression
1 bi2.04 ⊢ ¬ a = 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙ ↔ a · ˙ b = 0 ˙ → ¬ a = 0 ˙ → b = 0 ˙
2 df-ne ⊢ a ≠ 0 ˙ ↔ ¬ a = 0 ˙
3 2 imbi1i ⊢ a ≠ 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙ ↔ ¬ a = 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙
4 df-or ⊢ a = 0 ˙ ∨ b = 0 ˙ ↔ ¬ a = 0 ˙ → b = 0 ˙
5 4 imbi2i ⊢ a · ˙ b = 0 ˙ → a = 0 ˙ ∨ b = 0 ˙ ↔ a · ˙ b = 0 ˙ → ¬ a = 0 ˙ → b = 0 ˙
6 1 3 5 3bitr4ri ⊢ a · ˙ b = 0 ˙ → a = 0 ˙ ∨ b = 0 ˙ ↔ a ≠ 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙
7 6 2ralbii ⊢ ∀ a ∈ B ∀ b ∈ B a · ˙ b = 0 ˙ → a = 0 ˙ ∨ b = 0 ˙ ↔ ∀ a ∈ B ∀ b ∈ B a ≠ 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙
8 r19.21v ⊢ ∀ b ∈ B a ≠ 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙ ↔ a ≠ 0 ˙ → ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙
9 8 ralbii ⊢ ∀ a ∈ B ∀ b ∈ B a ≠ 0 ˙ → a · ˙ b = 0 ˙ → b = 0 ˙ ↔ ∀ a ∈ B a ≠ 0 ˙ → ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙
10 raldifsnb ⊢ ∀ a ∈ B a ≠ 0 ˙ → ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙ ↔ ∀ a ∈ B ∖ 0 ˙ ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙
11 7 9 10 3bitri ⊢ ∀ a ∈ B ∀ b ∈ B a · ˙ b = 0 ˙ → a = 0 ˙ ∨ b = 0 ˙ ↔ ∀ a ∈ B ∖ 0 ˙ ∀ b ∈ B a · ˙ b = 0 ˙ → b = 0 ˙