Theorem rexcom4b 3489

Description: Specialized existential commutation lemma. (Contributed by Jeff Madsen, 1-Jun-2011.)

Hypothesis

Ref	Expression
rexcom4b.1	⊢ 𝐵 ∈ V

Assertion

Ref	Expression
rexcom4b	⊢ (∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝑥 = 𝐵) ↔ ∃𝑦 ∈ 𝐴 𝜑)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝑦   𝜑,𝑥

Allowed substitution hints:   𝜑(𝑦)   𝐴(𝑦)   𝐵(𝑦)

Proof of Theorem rexcom4b

Step	Hyp	Ref	Expression
1		rexcom4a 3298	. 2 ⊢ (∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝑥 = 𝐵) ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ ∃𝑥 𝑥 = 𝐵))
2		rexcom4b.1	. . . . 5 ⊢ 𝐵 ∈ V
3	2	isseti 3476	. . . 4 ⊢ ∃𝑥 𝑥 = 𝐵
4	3	biantru 539	. . 3 ⊢ (𝜑 ↔ (𝜑 ∧ ∃𝑥 𝑥 = 𝐵))
5	4	rexbii 3115	. 2 ⊢ (∃𝑦 ∈ 𝐴 𝜑 ↔ ∃𝑦 ∈ 𝐴 (𝜑 ∧ ∃𝑥 𝑥 = 𝐵))
6	1, 5	bitr4i 281	1 ⊢ (∃𝑥∃𝑦 ∈ 𝐴 (𝜑 ∧ 𝑥 = 𝐵) ↔ ∃𝑦 ∈ 𝐴 𝜑)

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2146  ∃wrex 3092  Vcvv 3458

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-8 2148  ax-11 2195

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-clel 2841  df-rex 3093

This theorem is used by:  islshpat  39832