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Theorem isdomn5 20620
Description: The equivalence between the right conjuncts in the right hand sides of isdomn 20615 and isdomn2 20621, 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 2929 . . . . 5 (𝑎0 ↔ ¬ 𝑎 = 0 )
32imbi1i 349 . . . 4 ((𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ (¬ 𝑎 = 0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )))
4 df-or 848 . . . . 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 3107 . 2 (∀𝑎𝐵𝑏𝐵 ((𝑎 · 𝑏) = 0 → (𝑎 = 0𝑏 = 0 )) ↔ ∀𝑎𝐵𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )))
8 r19.21v 3157 . . 3 (∀𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ (𝑎0 → ∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 )))
98ralbii 3078 . 2 (∀𝑎𝐵𝑏𝐵 (𝑎0 → ((𝑎 · 𝑏) = 0𝑏 = 0 )) ↔ ∀𝑎𝐵 (𝑎0 → ∀𝑏𝐵 ((𝑎 · 𝑏) = 0𝑏 = 0 )))
10 raldifsnb 4743 . 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 847   = wceq 1541  wne 2928  wral 3047  cdif 3894  {csn 4571  (class class class)co 7341
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1968  ax-7 2009  ax-8 2113  ax-9 2121  ax-ext 2703
This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-tru 1544  df-ex 1781  df-sb 2068  df-clab 2710  df-cleq 2723  df-clel 2806  df-ne 2929  df-nel 3033  df-ral 3048  df-v 3438  df-dif 3900  df-sn 4572
This theorem is referenced by:  isdomn2  20621  isdomn4  20626
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