Theorem rexssOLD 4012

Description: Obsolete version of rexss 4010 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 3930	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	pm4.71rd 570	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
3	2	anbi1d 640	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) ∧ 𝜑)))
4		anass 472	. . 3 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑)))
5	3, 4	bitrdi 289	. 2 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6	5	rexbidv2 3181	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399   ∈ wcel 2141  ∃wrex 3085   ⊆ wss 3904

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1799  df-clel 2836  df-rex 3086  df-ss 3921

This theorem is referenced by: (None)