Theorem hbre1 1239

Description: x is not free in ∃x ∈ Aφ.

Assertion

Ref	Expression
hbre1	⊢ (∃x ∈ A φ → ∀x∃x ∈ A φ)

Proof of Theorem hbre1

Step	Hyp	Ref	Expression
1		hbe1 709	. 2 ⊢ (∃x(x ∈ A ∧ φ) → ∀x∃x(x ∈ A ∧ φ))
2		df-rex 1206	. 2 ⊢ (∃x ∈ A φ ↔ ∃x(x ∈ A ∧ φ))
3	2	bial 695	. 2 ⊢ (∀x∃x ∈ A φ ↔ ∀x∃x(x ∈ A ∧ φ))
4	1, 2, 3	3imtr4 192	1 ⊢ (∃x ∈ A φ → ∀x∃x ∈ A φ)

Syntax hints:   → wi 2   ∧ wa 196  ∀wal 672  ∃wex 678   ∈ wcel 1092  ∃wrex 1202

This theorem is referenced by:  hbiu1 2012  onfr 2237  zfregcl 3446  scott0 3542  chcmh 5148  atom1d 5750

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-gen 677

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-rex 1206

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