Theorem nss 1550

Description: Negation of subclass relationship. Exercise 13 of [TakeutiZaring] p. 18.

Assertion

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

Distinct variable group(s):   x,A   x,B

Proof of Theorem nss

Step	Hyp	Ref	Expression
1		exnal 721	. 2 ⊢ (∃x ¬ (x ∈ A → x ∈ B) ↔ ¬ ∀x(x ∈ A → x ∈ B))
2		annim 206	. . 3 ⊢ ((x ∈ A ∧ ¬ x ∈ B) ↔ ¬ (x ∈ A → x ∈ B))
3	2	biex 733	. 2 ⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) ↔ ∃x ¬ (x ∈ A → x ∈ B))
4		dfss2 1497	. . 3 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
5	4	negbii 162	. 2 ⊢ (¬ A ⊆ B ↔ ¬ ∀x(x ∈ A → x ∈ B))
6	1, 3, 5	3bitr4r 159	1 ⊢ (¬ A ⊆ B ↔ ∃x(x ∈ A ∧ ¬ x ∈ B))

Syntax hints:  ¬ wn 1   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672  ∃wex 678   ∈ wcel 1092   ⊆ wss 1487

This theorem is referenced by:  psslinpr 3929  reclem2pr 3951  shne0 5372

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-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

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