Theorem eximal 1698

Description: A utility theorem. An interesting case is when the same formula is substituted for both 𝜑 and 𝜓, since then both implications express a type of non-freeness. See also alimex 1748. (Contributed by BJ, 12-May-2019.)

Assertion

Ref	Expression
eximal	⊢ ((∃𝑥𝜑 → 𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))

Proof of Theorem eximal

Step	Hyp	Ref	Expression
1		df-ex 1696	. . 3 ⊢ (∃𝑥𝜑 ↔ ¬ ∀𝑥 ¬ 𝜑)
2	1	imbi1i 338	. 2 ⊢ ((∃𝑥𝜑 → 𝜓) ↔ (¬ ∀𝑥 ¬ 𝜑 → 𝜓))
3		con1b 347	. 2 ⊢ ((¬ ∀𝑥 ¬ 𝜑 → 𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))
4	2, 3	bitri 263	1 ⊢ ((∃𝑥𝜑 → 𝜓) ↔ (¬ 𝜓 → ∀𝑥 ¬ 𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 195  ∀wal 1473  ∃wex 1695

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

This theorem depends on definitions:  df-bi 196  df-ex 1696

This theorem is referenced by:  ax5e  1829  19.23t  2066  19.23tOLD  2206  xfree2  28688  bj-nalnalimiOLD  31799  bj-exalimi  31801