Theorem rexssOLD 4015

Description: Obsolete version of rexss 4013 as of 14-Oct-2025. (Contributed by Stefan O'Rear, 3-Apr-2015.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
rexssOLD	⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem rexssOLD

Step	Hyp	Ref	Expression
1		ssel 3933	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	pm4.71rd 571	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
3	2	anbi1d 642	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) ∧ 𝜑)))
4		anass 473	. . 3 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑)))
5	3, 4	bitrdi 290	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6	5	rexbidv2 3185	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 400   ∈ wcel 2145  ∃wrex 3089   ⊆ wss 3907

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1818  ax-4 1832  ax-5 1933  ax-6 1990  ax-7 2031  ax-8 2147

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1803  df-clel 2840  df-rex 3090  df-ss 3924

This theorem is referenced by: (None)