Theorem ralss 4004

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 3916	. . . 4 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		id 23	. . . . . . . 8 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	2	pm4.71rd 572	. . . . . . 7 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → (𝑥 ∈ 𝐴 ↔ (𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴)))
4	3	imbi1d 344	. . . . . 6 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ ((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑)))
5		impexp 456	. . . . . 6 ⊢ (((𝑥 ∈ 𝐵 ∧ 𝑥 ∈ 𝐴) → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
6	4, 5	bitrdi 290	. . . . 5 ⊢ ((𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
7	6	alimi 1844	. . . 4 ⊢ (∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵) → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
8	1, 7	sylbi 220	. . 3 ⊢ (𝐴 ⊆ 𝐵 → ∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
9		albi 1851	. . 3 ⊢ (∀𝑥((𝑥 ∈ 𝐴 → 𝜑) ↔ (𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))) → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
10	8, 9	syl 18	. 2 ⊢ (𝐴 ⊆ 𝐵 → (∀𝑥(𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑))))
11		df-ral 3078	. 2 ⊢ (∀𝑥 ∈ 𝐴 𝜑 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝜑))
12		df-ral 3078	. 2 ⊢ (∀𝑥 ∈ 𝐵 (𝑥 ∈ 𝐴 → 𝜑) ↔ ∀𝑥(𝑥 ∈ 𝐵 → (𝑥 ∈ 𝐴 → 𝜑)))
13	10, 11, 12	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 1568   ∈ wcel 2145  ∀wral 3077   ⊆ wss 3899

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842

This proof depends on definitions:  df-bi 210  df-an 402  df-ral 3078  df-ss 3916

This theorem is used by:  acsfn  17833  acsfn1  17835  acsfn2  17837  acsfn1p  21056  mdetunilem9  22935  ntrneik3  45095  ntrneix3  45096  ntrneik13  45097  ntrneix13  45098  ralabso  45957