Theorem ceqsex 3500

Description: Elimination of an existential quantifier, using implicit substitution. (Contributed by NM, 2-Mar-1995.) (Revised by Mario Carneiro, 10-Oct-2016.) (Proof shortened by Wolf Lammen, 22-Jan-2025.)

Hypotheses

Ref	Expression
ceqsex.1	⊢ Ⅎ𝑥𝜓
ceqsex.2	⊢ 𝐴 ∈ V
ceqsex.3	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ceqsex	⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓)

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑥)

Proof of Theorem ceqsex

Step	Hyp	Ref	Expression
1		alinexa 1876	. . 3 ⊢ (∀𝑥(𝑥 = 𝐴 → ¬ 𝜑) ↔ ¬ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
2		ceqsex.1	. . . . 5 ⊢ Ⅎ𝑥𝜓
3	2	nfn 1890	. . . 4 ⊢ Ⅎ𝑥 ¬ 𝜓
4		ceqsex.2	. . . 4 ⊢ 𝐴 ∈ V
5		ceqsex.3	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
6	5	notbid 321	. . . 4 ⊢ (𝑥 = 𝐴 → (¬ 𝜑 ↔ ¬ 𝜓))
7	3, 4, 6	ceqsal 3490	. . 3 ⊢ (∀𝑥(𝑥 = 𝐴 → ¬ 𝜑) ↔ ¬ 𝜓)
8	1, 7	bitr3i 280	. 2 ⊢ (¬ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ ¬ 𝜓)
9	8	con4bii 324	1 ⊢ (∃𝑥(𝑥 = 𝐴 ∧ 𝜑) ↔ 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568   = wceq 1570  ∃wex 1812  Ⅎwnf 1816   ∈ wcel 2145  Vcvv 3453

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-12 2215

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-clel 2837

This theorem is used by:  ceqsex2  3503