Theorem ralss 4058

Description: Restricted universal 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
ralss	⊢ (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 ralss

Step	Hyp	Ref	Expression
1		df-ss 3968	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		id 22	. . . . . . . 8 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 562	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
4	3	imbi1d 341	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑)))
5		impexp 450	. . . . . 6 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
6	4, 5	bitrdi 287	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
7	6	alimi 1811	. . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
8	1, 7	sylbi 217	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
9		albi 1818	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))) → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
10	8, 9	syl 17	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
11		df-ral 3062	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
12		df-ral 3062	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
13	10, 11, 12	3bitr4g 314	1 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1538   ∈ wcel 2108  ∀wral 3061   ⊆ wss 3951

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809

This theorem depends on definitions:  df-bi 207  df-an 396  df-ral 3062  df-ss 3968

This theorem is referenced by:  acsfn  17702  acsfn1  17704  acsfn2  17706  acsfn1p  20800  mdetunilem9  22626  ntrneik3  44109  ntrneix3  44110  ntrneik13  44111  ntrneix13  44112  ralabso  44985