Theorem r2exlem 3152

Description: Lemma factoring out common proof steps in r2exf 3285 an r2ex 3200. 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 1860	. . 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 338	. 2 ⊢ (∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑) ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
4		exnalimn 1877	. . 3 ⊢ (∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
5	4	exbii 1881	. 2 ⊢ (∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑) ↔ ∃𝑥 ¬ ∀𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) → ¬ 𝜑))
6		ralnex2 3143	. . 3 ⊢ (∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑 ↔ ¬ ∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑)
7	6	con2bii 360	. 2 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ¬ ∀𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 ¬ 𝜑)
8	3, 5, 7	3bitr4ri 307	1 ⊢ (∃𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 𝜑 ↔ ∃𝑥∃𝑦((𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐵) ∧ 𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812   ∈ wcel 2145  ∀wral 3077  ∃wrex 3087

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-ex 1813  df-ral 3078  df-rex 3088

This theorem is used by:  r2ex  3200  r2exf  3285