Theorem rexss 4010

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 3921	. . 3 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		pm3.41 496	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 570	. . . 4 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
4	3	alexbii 1852	. . 3 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
5	1, 4	sylbi 219	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑))))
6		df-rex 3086	. 2 ⊢ (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
7		df-rex 3086	. 2 ⊢ (∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑥(𝑥 ∈ 𝐵 ∧ (𝑥 ∈ 𝐴 ∧ 𝜑)))
8	5, 6, 7	3bitr4g 316	1 ⊢ (𝐴 ⊆ 𝐵 → (∃𝑥 ∈ 𝐴 𝜑 ↔ ∃𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 ∧ 𝜑)))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 399  ∀wal 1557  ∃wex 1798   ∈ 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

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

This theorem is referenced by:  oddnn02np1  16365  oddge22np1  16366  evennn02n  16367  evennn2n  16368  2lgslem1a  27432  omssubadd  34558  rexabso  45509  limsupmnfuzlem  46264  sbgoldbo  48373