Theorem hbreu1 1307

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

Assertion

Ref	Expression
hbreu1	⊢ (∃!x ∈ A φ → ∀x∃!x ∈ A φ)

Proof of Theorem hbreu1

Step	Hyp	Ref	Expression
1		hbeu1 1015	. 2 ⊢ (∃!x(x ∈ A ∧ φ) → ∀x∃!x(x ∈ A ∧ φ))
2		df-reu 1207	. 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  ∃!weu 1007   ∈ wcel 1092  ∃!wreu 1203

This theorem is referenced by:  reuuni2 1956  reuuni4 1959

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-7 676  ax-gen 677

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-eu 1009  df-reu 1207

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