Theorem nssss 1866

Description: Negation of subclass relationship. Compare nss 1550.

Assertion

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

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

Proof of Theorem nssss

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		ssextss 1864	. . 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   ⊆ wss 1487

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-13 804  ax-14 805  ax-16 922  ax-17 925  ax-ext 1074  ax-rep 1075  ax-pow 1077

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-dif 1489  df-in 1491  df-ss 1492  df-nul 1708  df-pw 1799  df-sn 1811

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