Theorem rexss 3982

Description: Restricted existential quantification on a subset in terms of superset. (Contributed by Stefan O'Rear, 3-Apr-2015.) Avoid axioms. (Revised by SN, 14-Oct-2025.)

Assertion

Ref	Expression
rexss	⊢ (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 rexss

Step	Hyp	Ref	Expression
1		df-ss 3896	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		pm3.41 498	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 572	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
4	3	alexbii 1865	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	1, 4	sylbi 220	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6		df-rex 3087	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
7		df-rex 3087	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑)))
8	5, 6, 7	3bitr4g 317	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567  ∃wex 1811   ∈ wcel 2144  ∃wrex 3086   ⊆ wss 3895

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1812  df-rex 3087  df-ss 3896

This theorem is used by:  oddnn02np1  16524  oddge22np1  16525  evennn02n  16526  evennn2n  16527  2lgslem1a  27724  omssubadd  34979  rexabso  46053  limsupmnfuzlem  46830  sbgoldbo  48979