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Theorem isdomn5 20655
Description: The equivalence between the right conjuncts in the right hand sides of isdomn 20650 and isdomn2 20656, in predicate calculus form. (Contributed by SN, 16-Sep-2024.)
Assertion
Ref Expression
isdomn5 (∀𝑎𝐵𝑏𝐵 ((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ ∀𝑎 ∈ (𝐵 ∖ { 0 })∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 ))
Distinct variable group:   0 ,𝑎,𝑏
Allowed substitution hints:   𝐵(𝑎,𝑏)   · (𝑎,𝑏)

Proof of Theorem isdomn5
StepHypRef Expression
1 bi2.04 387 . . . 4 ((¬ 𝑎 = 0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ ((𝑎 · 𝑏) = 0 → (¬ 𝑎 = 0𝑏 = 0 )))
2 df-ne 2934 . . . . 5 (𝑎0 ↔ ¬ 𝑎 = 0 )
32imbi1i 349 . . . 4 ((𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ (¬ 𝑎 = 0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )))
4 df-or 849 . . . . 5 ((𝑎 = 0𝑏 = 0 ) ↔ (¬ 𝑎 = 0𝑏 = 0 ))
54imbi2i 336 . . . 4 (((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ ((𝑎 · 𝑏) = 0 → (¬ 𝑎 = 0𝑏 = 0 )))
61, 3, 53bitr4ri 304 . . 3 (((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )))
762ralbii 3113 . 2 (∀𝑎𝐵𝑏𝐵 ((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ ∀𝑎𝐵𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )))
8 r19.21v 3163 . . 3 (∀𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ (𝑎0 → ∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 )))
98ralbii 3084 . 2 (∀𝑎𝐵𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ ∀𝑎𝐵 (𝑎0 → ∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 )))
10 raldifsnb 4754 . 2 (∀𝑎𝐵 (𝑎0 → ∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ ∀𝑎 ∈ (𝐵 ∖ { 0 })∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 ))
117, 9, 103bitri 297 1 (∀𝑎𝐵𝑏𝐵 ((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ ∀𝑎 ∈ (𝐵 ∖ { 0 })∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 ))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wi 4  wb 206  wo 848   = wceq 1542  wne 2933  wral 3052  cdif 3900  {csn 4582  (class class class)co 7368
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1912  ax-6 1969  ax-7 2010  ax-8 2116  ax-9 2124  ax-ext 2709
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 849  df-tru 1545  df-ex 1782  df-sb 2069  df-clab 2716  df-cleq 2729  df-clel 2812  df-ne 2934  df-nel 3038  df-ral 3053  df-v 3444  df-dif 3906  df-sn 4583
This theorem is referenced by:  isdomn2  20656  isdomn4  20661
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