Theorem bj-biexal3 32698

Description: A general FOL biconditional. (Contributed by BJ, 20-Oct-2019.)

Assertion

Ref	Expression
bj-biexal3	⊢ (∀𝑥(𝜑 → ∀𝑥𝜓) ↔ ∀𝑥(∃𝑥𝜑 → 𝜓))

Proof of Theorem bj-biexal3

Step	Hyp	Ref	Expression
1		bj-biexal1 32696	. 2 ⊢ (∀𝑥(𝜑 → ∀𝑥𝜓) ↔ (∃𝑥𝜑 → ∀𝑥𝜓))
2		bj-biexal2 32697	. 2 ⊢ (∀𝑥(∃𝑥𝜑 → 𝜓) ↔ (∃𝑥𝜑 → ∀𝑥𝜓))
3	1, 2	bitr4i 267	1 ⊢ (∀𝑥(𝜑 → ∀𝑥𝜓) ↔ ∀𝑥(∃𝑥𝜑 → 𝜓))

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1481  ∃wex 1704

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047

This theorem depends on definitions:  df-bi 197  df-or 385  df-ex 1705  df-nf 1710

This theorem is referenced by: (None)