Theorem unssd 3440

Description: A deduction showing the union of two subclasses is a subclass. (Contributed by Jonathan Ben-Naim, 3-Jun-2011.)

Hypotheses

Ref	Expression
unssd.1	⊢ (φ → A ⊆ C)
unssd.2	⊢ (φ → B ⊆ C)

Assertion

Ref	Expression
unssd	⊢ (φ → (A ∪ B) ⊆ C)

Proof of Theorem unssd

Step	Hyp	Ref	Expression
1		unssd.1	. 2 ⊢ (φ → A ⊆ C)
2		unssd.2	. 2 ⊢ (φ → B ⊆ C)
3		unss 3438	. . 3 ⊢ ((A ⊆ C ∧ B ⊆ C) ↔ (A ∪ B) ⊆ C)
4	3	biimpi 186	. 2 ⊢ ((A ⊆ C ∧ B ⊆ C) → (A ∪ B) ⊆ C)
5	1, 2, 4	syl2anc 642	1 ⊢ (φ → (A ∪ B) ⊆ C)

This proof depends on syntax axioms:   → wi 4   ∧ wa 358   ∪ cun 3208   ⊆ wss 3258

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-ss 3260

This theorem is used by:  nchoicelem6  6295  frecsuc  6323