Theorem exsb 2390

Description: An equivalent expression for existence. One direction (exsbim 2035) needs fewer axioms. (Contributed by NM, 2-Feb-2005.) Avoid ax-13 2403. (Revised by Wolf Lammen, 16-Oct-2022.)

Assertion

Ref	Expression
exsb	⊢ (∃𝑥𝜑 ↔ ∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑))

Distinct variable groups:   𝑥,𝑦   𝜑,𝑦

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem exsb

Step	Hyp	Ref	Expression
1		nfv 1947	. 2 ⊢ Ⅎ𝑦𝜑
2		nfa1 2188	. 2 ⊢ Ⅎ𝑥∀𝑥(𝑥 = 𝑦 → 𝜑)
3		ax12v 2216	. . 3 ⊢ (𝑥 = 𝑦 → (𝜑 → ∀𝑥(𝑥 = 𝑦 → 𝜑)))
4		sp 2221	. . . 4 ⊢ (∀𝑥(𝑥 = 𝑦 → 𝜑) → (𝑥 = 𝑦 → 𝜑))
5	4	com12 33	. . 3 ⊢ (𝑥 = 𝑦 → (∀𝑥(𝑥 = 𝑦 → 𝜑) → 𝜑))
6	3, 5	impbid 215	. 2 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ ∀𝑥(𝑥 = 𝑦 → 𝜑)))
7	1, 2, 6	cbvexv1 2373	1 ⊢ (∃𝑥𝜑 ↔ ∃𝑦∀𝑥(𝑥 = 𝑦 → 𝜑))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀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 2178  ax-11 2194  ax-12 2215

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)