Theorem nrexralim 3225

Description: Negation of a complex predicate calculus formula. (Contributed by FL, 31-Jul-2009.)

Assertion

Ref	Expression
nrexralim	⊢ (¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓))

Proof of Theorem nrexralim

Step	Hyp	Ref	Expression
1		rexanali 3224	. . 3 ⊢ (∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓) ↔ ¬ ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓))
2	1	ralbii 3133	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ¬ ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓))
3		ralnex 3199	. 2 ⊢ (∀𝑥 ∈ 𝐴 ¬ ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ↔ ¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓))
4	2, 3	bitr2i 279	1 ⊢ (¬ ∃𝑥 ∈ 𝐴 ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓) ↔ ∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 399  ∀wral 3106  ∃wrex 3107

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811

This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-ral 3111  df-rex 3112

This theorem is referenced by: (None)