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Theorem 2reu5lem2 3713
Description: Lemma for 2reu5 3715. (Contributed by Alexander van der Vekens, 17-Jun-2017.)
Assertion
Ref Expression
2reu5lem2 (∀𝑥 ∈ 𝐴 ∃*𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
Distinct variable groups:   𝑦,𝐴   𝑥,𝑦
Allowed substitution hints:   𝜑(𝑥, 𝑦)   𝐴(𝑥)   𝐵(𝑥, 𝑦)

Proof of Theorem 2reu5lem2
StepHypRef Expression
1 df-rmo 3365 . . 3 (∃*𝑦 ∈ 𝐵 𝜑 ↔ ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
21ralbii 3108 . 2 (∀𝑥 ∈ 𝐴 ∃*𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥 ∈ 𝐴 ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑))
3 df-ral 3077 . . 3 (∀𝑥 ∈ 𝐴 ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐴 → ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
4 moanimv 2644 . . . . . 6 (∃*𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 → ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)))
54bicomi 227 . . . . 5 ((𝑥 ∈ 𝐴 → ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃*𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
6 3anass 1111 . . . . . . 7 ((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑) ↔ (𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)))
76bicomi 227 . . . . . 6 ((𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ (𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
87mobii 2573 . . . . 5 (∃*𝑦(𝑥 ∈ 𝐴 ∧ (𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
95, 8bitri 278 . . . 4 ((𝑥 ∈ 𝐴 → ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
109albii 1851 . . 3 (∀𝑥(𝑥 ∈ 𝐴 → ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑)) ↔ ∀𝑥∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
113, 10bitri 278 . 2 (∀𝑥 ∈ 𝐴 ∃*𝑦(𝑦 ∈ 𝐵 ∧ 𝜑) ↔ ∀𝑥∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
122, 11bitri 278 1 (∀𝑥 ∈ 𝐴 ∃*𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥∃*𝑦(𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵 ∧ 𝜑))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103  ∀wal 1567   ∈ wcel 2144  ∃*wmo 2562  ∀wral 3076  ∃*wrmo 3364
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040
This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-ex 1812  df-mo 2564  df-ral 3077  df-rmo 3365
This theorem is used by:  2reu5lem3  3714
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