Theorem unss 1632

Description: The union of two subclasses is a subclass. Theorem 27 of [Suppes] p. 27 and its converse.

Assertion

Ref	Expression
unss	⊢ ((A ⊆ C ∧ B ⊆ C) ↔ (A ∪ B) ⊆ C)

Proof of Theorem unss

Step	Hyp	Ref	Expression
1		elun 1601	. . . . 5 ⊢ (x ∈ (A ∪ B) ↔ (x ∈ A ∨ x ∈ B))
2	1	imbi1i 161	. . . 4 ⊢ ((x ∈ (A ∪ B) → x ∈ C) ↔ ((x ∈ A ∨ x ∈ B) → x ∈ C))
3		jaob 328	. . . 4 ⊢ (((x ∈ A ∨ x ∈ B) → x ∈ C) ↔ ((x ∈ A → x ∈ C) ∧ (x ∈ B → x ∈ C)))
4	2, 3	bitr 151	. . 3 ⊢ ((x ∈ (A ∪ B) → x ∈ C) ↔ ((x ∈ A → x ∈ C) ∧ (x ∈ B → x ∈ C)))
5	4	bial 695	. 2 ⊢ (∀x(x ∈ (A ∪ B) → x ∈ C) ↔ ∀x((x ∈ A → x ∈ C) ∧ (x ∈ B → x ∈ C)))
6		dfss2 1497	. 2 ⊢ ((A ∪ B) ⊆ C ↔ ∀x(x ∈ (A ∪ B) → x ∈ C))
7		dfss2 1497	. . . 4 ⊢ (A ⊆ C ↔ ∀x(x ∈ A → x ∈ C))
8		dfss2 1497	. . . 4 ⊢ (B ⊆ C ↔ ∀x(x ∈ B → x ∈ C))
9	7, 8	anbi12i 369	. . 3 ⊢ ((A ⊆ C ∧ B ⊆ C) ↔ (∀x(x ∈ A → x ∈ C) ∧ ∀x(x ∈ B → x ∈ C)))
10		19.26 749	. . 3 ⊢ (∀x((x ∈ A → x ∈ C) ∧ (x ∈ B → x ∈ C)) ↔ (∀x(x ∈ A → x ∈ C) ∧ ∀x(x ∈ B → x ∈ C)))
11	9, 10	bitr4 154	. 2 ⊢ ((A ⊆ C ∧ B ⊆ C) ↔ ∀x((x ∈ A → x ∈ C) ∧ (x ∈ B → x ∈ C)))
12	5, 6, 11	3bitr4r 159	1 ⊢ ((A ⊆ C ∧ B ⊆ C) ↔ (A ∪ B) ⊆ C)

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   ∧ wa 196  ∀wal 672   ∈ wcel 1092   ∪ cun 1485   ⊆ wss 1487

This theorem is referenced by:  unssi 1633  nsspssun 1666  suceloni 2314  xpex 2488  relun 2490  er2 3201  trcl 3489  infxpidmlem11 4943  dfchj2 5325  sshjclt 5328  shlub 5347  spanun 5450

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-in 1491  df-ss 1492

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