Theorem r2exlem 3267

Description: Lemma factoring out common proof steps in r2exf 3288 an r2ex 3268. 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 1812	. . 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 336	. 2 ⊢ (∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑) ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
4		exnalimn 1829	. . 3 ⊢ (∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
5	4	exbii 1833	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
6		ralnex2 3226	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑)
7	6	con2bii 359	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
8	3, 5, 7	3bitr4ri 305	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 207   ∧ wa 396  ∀wal 1523  ∃wex 1765   ∈ wcel 2083  ∀wral 3107  ∃wrex 3108

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1781  ax-4 1795

This theorem depends on definitions:  df-bi 208  df-an 397  df-ex 1766  df-ral 3112  df-rex 3113

This theorem is referenced by:  r2ex  3268  r2exf  3288