Theorem nss 3330

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18. (Contributed by NM, 25-Feb-1996.) (Proof shortened by Andrew Salmon, 21-Jun-2011.)

Assertion

Ref	Expression
nss	⊢ (¬ A ⊆ B ↔ ∃x(x ∈ A ∧ ¬ x ∈ B))

Distinct variable groups:   x,A   x,B

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exanali 1585	. . 3 ⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) ↔ ¬ ∀x(x ∈ A → x ∈ B))
2		dfss2 3263	. . 3 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
3	1, 2	xchbinxr 302	. 2 ⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) ↔ ¬ A ⊆ B)
4	3	bicomi 193	1 ⊢ (¬ A ⊆ B ↔ ∃x(x ∈ A ∧ ¬ x ∈ B))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ wa 358  ∀wal 1540  ∃wex 1541   ∈ wcel 1710   ⊆ 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-ss 3260

This theorem is used by: (None)