Theorem intmin4 3916

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 3905	. . . 4 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝐴 ⊆ 𝑥))
2		simpr 110	. . . . . . . 8 ⊢ ((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝜑)
3		ancr 321	. . . . . . . 8 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (𝜑 → (𝐴 ⊆ 𝑥 ∧ 𝜑)))
4	2, 3	impbid2 143	. . . . . . 7 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → ((𝐴 ⊆ 𝑥 ∧ 𝜑) ↔ 𝜑))
5	4	imbi1d 231	. . . . . 6 ⊢ ((𝜑 → 𝐴 ⊆ 𝑥) → (((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
6	5	alimi 1479	. . . . 5 ⊢ (∀𝑥(𝜑 → 𝐴 ⊆ 𝑥) → ∀𝑥(((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ (𝜑 → 𝑦 ∈ 𝑥)))
7		albi 1492	. . . . 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 121	. . 3 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥) ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥)))
10		vex 2776	. . . 4 ⊢ 𝑦 ∈ V
11	10	elintab 3899	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ ∀𝑥((𝐴 ⊆ 𝑥 ∧ 𝜑) → 𝑦 ∈ 𝑥))
12	10	elintab 3899	. . 3 ⊢ (𝑦 ∈ ∩ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝜑 → 𝑦 ∈ 𝑥))
13	9, 11, 12	3bitr4g 223	. 2 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → (𝑦 ∈ ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} ↔ 𝑦 ∈ ∩ {𝑥 ∣ 𝜑}))
14	13	eqrdv 2204	1 ⊢ (𝐴 ⊆ ∩ {𝑥 ∣ 𝜑} → ∩ {𝑥 ∣ (𝐴 ⊆ 𝑥 ∧ 𝜑)} = ∩ {𝑥 ∣ 𝜑})

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1371   = wceq 1373   ∈ wcel 2177  {cab 2192   ⊆ wss 3168  ∩ cint 3888

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-ral 2490  df-v 2775  df-in 3174  df-ss 3181  df-int 3889

This theorem is referenced by: (None)