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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 ) )