Theorem rexssOLD 4016

Description: Obsolete version of rexss 4014 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 3934	. . . . 5 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2	1	pm4.71rd 572	. . . 4 ⊢ (𝐴 ⊆ 𝐵 → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
3	2	anbi1d 643	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) ∧ 𝜑)))
4		anass 474	. . 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 3188	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∈ wcel 2146  ∃wrex 3092   ⊆ wss 3908

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

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-clel 2841  df-rex 3093  df-ss 3925

This theorem is used by: (None)