Theorem intmin4 3860

Description: Elimination of a conjunct in a class intersection. (Contributed by NM, 31-Jul-2006.)

Assertion

Ref	Expression
intmin4	⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} = ∩ {𝑥 ∣ 𝜑})

Distinct variable group:   𝑥,𝐴

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem intmin4

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		ssintab 3849	. . . 4 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝐴 ⊆ 𝑥))
2		simpr 109	. . . . . . . 8 ⊢ ((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝜑)
3		ancr 319	. . . . . . . 8 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (𝜑 → (𝐴 ⊆ 𝑥 ∧ 𝜑)))
4	2, 3	impbid2 142	. . . . . . 7 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → ((𝐴 ⊆ 𝑥 ∧ 𝜑) ↔ 𝜑))
5	4	imbi1d 230	. . . . . 6 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
6	5	alimi 1449	. . . . 5 ⊢ (∀𝑥(𝜑 → 𝐴 ⊆ 𝑥) → ∀𝑥(((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
7		albi 1462	. . . . 5 ⊢ (∀𝑥(((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)) → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
8	6, 7	syl 14	. . . 4 ⊢ (∀𝑥(𝜑 → 𝐴 ⊆ 𝑥) → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
9	1, 8	sylbi 120	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
10		vex 2734	. . . 4 ⊢ 𝑦 ∈ V
11	10	elintab 3843	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ ∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥))
12	10	elintab 3843	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥))
13	9, 11, 12	3bitr4g 222	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ 𝑦 ∈ ∩ {𝑥 ∣ 𝜑}))
14	13	eqrdv 2169	1 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} = ∩ {𝑥 ∣ 𝜑})

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1347   = wceq 1349   ∈ wcel 2142  {cab 2157   ⊆ wss 3122  ∩ cint 3832

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 705  ax-5 1441  ax-7 1442  ax-gen 1443  ax-ie1 1487  ax-ie2 1488  ax-8 1498  ax-10 1499  ax-11 1500  ax-i12 1501  ax-bndl 1503  ax-4 1504  ax-17 1520  ax-i9 1524  ax-ial 1528  ax-i5r 1529  ax-ext 2153

This theorem depends on definitions:  df-bi 116  df-tru 1352  df-nf 1455  df-sb 1757  df-clab 2158  df-cleq 2164  df-clel 2167  df-nfc 2302  df-ral 2454  df-v 2733  df-in 3128  df-ss 3135  df-int 3833

This theorem is referenced by: (None)