Theorem birex2 1227

Description: Inference adding different restricted existential quantifiers to each side of an equivalence.

Hypothesis

Ref	Expression
birex2.1	⊢ ((x ∈ A ∧ φ) ↔ (x ∈ B ∧ ψ))

Assertion

Ref	Expression
birex2	⊢ (∃x ∈ A φ ↔ ∃x ∈ B ψ)

Proof of Theorem birex2

Step	Hyp	Ref	Expression
1		birex2.1	. . 3 ⊢ ((x ∈ A ∧ φ) ↔ (x ∈ B ∧ ψ))
2	1	biex 733	. 2 ⊢ (∃x(x ∈ A ∧ φ) ↔ ∃x(x ∈ B ∧ ψ))
3		df-rex 1206	. 2 ⊢ (∃x ∈ A φ ↔ ∃x(x ∈ A ∧ φ))
4		df-rex 1206	. 2 ⊢ (∃x ∈ B ψ ↔ ∃x(x ∈ B ∧ ψ))
5	2, 3, 4	3bitr4 158	1 ⊢ (∃x ∈ A φ ↔ ∃x ∈ B ψ)

Syntax hints:   ↔ wb 127   ∧ wa 196  ∃wex 678   ∈ wcel 1092  ∃wrex 1202

This theorem is referenced by:  birexa 1229  wefrc 2195  bnd2 3549  sumdmdi 5785

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-gen 677

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

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