Theorem nrexralim 3147

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 3117	. . 3 ⊢ (∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓) ↔ ¬ ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓))
2	1	ralbii 3109	. 2 ⊢ (∀𝑥 ∈ 𝐴 ∃𝑦 ∈ 𝐵 (𝜑 ∧ ¬ 𝜓) ↔ ∀𝑥 ∈ 𝐴 ¬ ∀𝑦 ∈ 𝐵 (𝜑 → 𝜓))
3		ralnex 3089	. 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))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀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: (None)