Theorem r2exlem 3133

Description: Lemma factoring out common proof steps in r2exf 3270 an r2ex 3186. Introduced to reduce dependencies on axioms. (Contributed by Wolf Lammen, 10-Jan-2020.)

Hypothesis

Ref	Expression
r2exlem.1	⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))

Assertion

Ref	Expression
r2exlem	⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))

Proof of Theorem r2exlem

Step	Hyp	Ref	Expression
1		exnal 1822	. . 3 ⊢ (∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑) ↔ ¬ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
2		r2exlem.1	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ∀𝑥∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
3	1, 2	xchbinxr 334	. 2 ⊢ (∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑) ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
4		exnalimn 1839	. . 3 ⊢ (∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
5	4	exbii 1843	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
6		ralnex2 3123	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑)
7	6	con2bii 356	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
8	3, 5, 7	3bitr4ri 303	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 394  ∀wal 1532  ∃wex 1774   ∈ wcel 2099  ∀wral 3051  ∃wrex 3060

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804

This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1775  df-ral 3052  df-rex 3061

This theorem is referenced by:  r2ex  3186  r2exf  3270