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Theorem r2allem 3151
Description: Lemma factoring out common proof steps of r2alf 3284 and r2al 3199. Introduced to reduce dependencies on axioms. (Contributed by Wolf Lammen, 9-Jan-2020.)
Hypothesis
Ref Expression
r2allem.1 (∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))
Assertion
Ref Expression
r2allem (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑))

Proof of Theorem r2allem
StepHypRef Expression
1 df-ral 3078 . 2 (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
2 r2allem.1 . . . 4 (∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))
3 impexp 456 . . . . 5 (((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ (𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)))
43albii 1852 . . . 4 (∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ ∀𝑦(𝑥 ∈ 𝐴 → (𝑦 ∈ 𝐵 → 𝜑)))
5 df-ral 3078 . . . . 5 (∀𝑦 ∈ 𝐵 𝜑 ↔ ∀𝑦(𝑦 ∈ 𝐵 → 𝜑))
65imbi2i 339 . . . 4 ((𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑) ↔ (𝑥 ∈ 𝐴 → ∀𝑦(𝑦 ∈ 𝐵 → 𝜑)))
72, 4, 63bitr4i 306 . . 3 (∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ (𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
87albii 1852 . 2 (∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐴 → ∀𝑦 ∈ 𝐵 𝜑))
91, 8bitr4i 281 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  ∀wal 1568   ∈ wcel 2145  ∀wral 3077
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842
This proof depends on definitions:  df-bi 210  df-an 402  df-ral 3078
This theorem is used by:  r2al  3199  r2alf  3284
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