Theorem alexbii 1865

Description: Biconditional form of aleximi 1864. (Contributed by BJ, 16-Nov-2020.)

Hypothesis

Ref	Expression
alexbii.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
alexbii	⊢ (∀𝑥𝜑 → (∃𝑥𝜓 ↔ ∃𝑥𝜒))

Proof of Theorem alexbii

Step	Hyp	Ref	Expression
1		alexbii.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	biimpd 232	. . 3 ⊢ (𝜑 → (𝜓 → 𝜒))
3	2	aleximi 1864	. 2 ⊢ (∀𝑥𝜑 → (∃𝑥𝜓 → ∃𝑥𝜒))
4	1	biimprd 251	. . 3 ⊢ (𝜑 → (𝜒 → 𝜓))
5	4	aleximi 1864	. 2 ⊢ (∀𝑥𝜑 → (∃𝑥𝜒 → ∃𝑥𝜓))
6	3, 5	impbid 215	1 ⊢ (∀𝑥𝜑 → (∃𝑥𝜓 ↔ ∃𝑥𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567  ∃wex 1811

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841

This proof depends on definitions:  df-bi 210  df-ex 1812

This theorem is used by:  exbi  1879  exbidh  1899  exintrbi  1923  eleq2d  2846  rexeq  3315  rexss  3983  ttrclselem2  9734  bnj956  35457  bj-2exbi  37625  bj-axreprepsep  38127