Theorem r2exlem 3143

Description: Lemma factoring out common proof steps in r2exf 3282 an r2ex 3196. 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 1827	. . 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 335	. 2 ⊢ (∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑) ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
4		exnalimn 1844	. . 3 ⊢ (∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
5	4	exbii 1848	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
6		ralnex2 3133	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑)
7	6	con2bii 357	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
8	3, 5, 7	3bitr4ri 304	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1538  ∃wex 1779   ∈ wcel 2108  ∀wral 3061  ∃wrex 3070

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-ral 3062  df-rex 3071

This theorem is referenced by:  r2ex  3196  r2exf  3282