Theorem r19.2zb 3641

Description: A response to the notion that the condition A ≠ ∅ can be removed in r19.2z 3640. Interestingly enough, φ does not figure in the left-hand side. (Contributed by Jeff Hankins, 24-Aug-2009.)

Assertion

Ref	Expression
r19.2zb	⊢ (A ≠ ∅ ↔ (∀x ∈ A φ → ∃x ∈ A φ))

Distinct variable group:   x,A

Allowed substitution hint:   φ(x)

Proof of Theorem r19.2zb

Step	Hyp	Ref	Expression
1		r19.2z 3640	. . 3 ⊢ ((A ≠ ∅ ∧ ∀x ∈ A φ) → ∃x ∈ A φ)
2	1	ex 423	. 2 ⊢ (A ≠ ∅ → (∀x ∈ A φ → ∃x ∈ A φ))
3		noel 3555	. . . . . . 7 ⊢ ¬ x ∈ ∅
4	3	pm2.21i 123	. . . . . 6 ⊢ (x ∈ ∅ → φ)
5	4	rgen 2680	. . . . 5 ⊢ ∀x ∈ ∅ φ
6		raleq 2808	. . . . 5 ⊢ (A = ∅ → (∀x ∈ A φ ↔ ∀x ∈ ∅ φ))
7	5, 6	mpbiri 224	. . . 4 ⊢ (A = ∅ → ∀x ∈ A φ)
8	7	necon3bi 2558	. . 3 ⊢ (¬ ∀x ∈ A φ → A ≠ ∅)
9		exsimpl 1592	. . . 4 ⊢ (∃x(x ∈ A ∧ φ) → ∃x x ∈ A)
10		df-rex 2621	. . . 4 ⊢ (∃x ∈ A φ ↔ ∃x(x ∈ A ∧ φ))
11		n0 3560	. . . 4 ⊢ (A ≠ ∅ ↔ ∃x x ∈ A)
12	9, 10, 11	3imtr4i 257	. . 3 ⊢ (∃x ∈ A φ → A ≠ ∅)
13	8, 12	ja 153	. 2 ⊢ ((∀x ∈ A φ → ∃x ∈ A φ) → A ≠ ∅)
14	2, 13	impbii 180	1 ⊢ (A ≠ ∅ ↔ (∀x ∈ A φ → ∃x ∈ A φ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ wa 358  ∃wex 1541   = wceq 1642   ∈ wcel 1710   ≠ wne 2517  ∀wral 2615  ∃wrex 2616  ∅c0 3551

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-ne 2519  df-ral 2620  df-rex 2621  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-dif 3216  df-nul 3552

This theorem is used by: (None)