Theorem ceqsal1t 3499

Description: One direction of ceqsalt 3500 is based on fewer assumptions and fewer axioms. It is at the same time the reverse direction of vtoclgft 3535. Extracted from a proof of ceqsalt 3500. (Contributed by Wolf Lammen, 25-Mar-2025.)

Assertion

Ref	Expression
ceqsal1t	⊢ ((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝜓 → ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Proof of Theorem ceqsal1t

Step	Hyp	Ref	Expression
1		biimpr 219	. . . . . 6 ⊢ ((𝜑 ↔ 𝜓) → (𝜓 → 𝜑))
2	1	imim2i 16	. . . . 5 ⊢ ((𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → (𝑥 = 𝐴 → (𝜓 → 𝜑)))
3	2	com23 86	. . . 4 ⊢ ((𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → (𝜓 → (𝑥 = 𝐴 → 𝜑)))
4	3	alimi 1805	. . 3 ⊢ (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → ∀𝑥(𝜓 → (𝑥 = 𝐴 → 𝜑)))
5		19.21t 2191	. . 3 ⊢ (Ⅎ𝑥𝜓 → (∀𝑥(𝜓 → (𝑥 = 𝐴 → 𝜑)) ↔ (𝜓 → ∀𝑥(𝑥 = 𝐴 → 𝜑))))
6	4, 5	imbitrid 243	. 2 ⊢ (Ⅎ𝑥𝜓 → (∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓)) → (𝜓 → ∀𝑥(𝑥 = 𝐴 → 𝜑))))
7	6	imp 406	1 ⊢ ((Ⅎ𝑥𝜓 ∧ ∀𝑥(𝑥 = 𝐴 → (𝜑 ↔ 𝜓))) → (𝜓 → ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395  ∀wal 1531   = wceq 1533  Ⅎwnf 1777

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-12 2163

This theorem depends on definitions:  df-bi 206  df-an 396  df-ex 1774  df-nf 1778

This theorem is referenced by:  ceqsalt  3500