Theorem sbalexOLD 2282

Description: Obsolete version of sbalex 2281 as of 14-Aug-2025. (Contributed by NM, 14-Apr-2008.) (Revised by BJ, 20-Dec-2020.) (Revised by BJ, 21-Sep-2024.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbalexOLD	⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

Distinct variable group:   𝑥,𝑡

Allowed substitution hints:   𝜑(𝑥, 𝑡)

Proof of Theorem sbalexOLD

Step	Hyp	Ref	Expression
1		nfa1 2189	. . 3 ⊢ Ⅎ𝑥∀𝑥(𝑥 = 𝑡 → 𝜑)
2		ax12v2 2218	. . . 4 ⊢ (𝑥 = 𝑡 → (𝜑 → ∀𝑥(𝑥 = 𝑡 → 𝜑)))
3	2	imp 412	. . 3 ⊢ ((𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
4	1, 3	exlimi 2256	. 2 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) → ∀𝑥(𝑥 = 𝑡 → 𝜑))
5		equs4v 2033	. 2 ⊢ (∀𝑥(𝑥 = 𝑡 → 𝜑) → ∃𝑥(𝑥 = 𝑡 ∧ 𝜑))
6	4, 5	impbii 212	1 ⊢ (∃𝑥(𝑥 = 𝑡 ∧ 𝜑) ↔ ∀𝑥(𝑥 = 𝑡 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1568  ∃wex 1812

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-10 2179  ax-12 2216

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

This theorem is used by: (None)