Theorem nssr 2980

Description: Negation of subclass relationship. One direction of Exercise 13 of [TakeutiZaring] p. 18. (Contributed by Jim Kingdon, 15-Jul-2018.)

Assertion

Ref	Expression
nssr	⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) → ¬ A ⊆ B)

Distinct variable groups:   x,A   x,B

Proof of Theorem nssr

Step	Hyp	Ref	Expression
1		exanaliim 1520	. 2 ⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) → ¬ ∀x(x ∈ A → x ∈ B))
2		dfss2 2911	. 2 ⊢ (A ⊆ B ↔ ∀x(x ∈ A → x ∈ B))
3	1, 2	sylnibr 589	1 ⊢ (∃x(x ∈ A ∧ ¬ x ∈ B) → ¬ A ⊆ B)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97  ∀wal 1226  ∃wex 1362   ∈ wcel 1374   ⊆ wss 2894

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-5 1316  ax-7 1317  ax-gen 1318  ax-ie1 1363  ax-ie2 1364  ax-8 1376  ax-11 1378  ax-4 1381  ax-17 1400  ax-i9 1404  ax-ial 1409  ax-i5r 1410  ax-ext 2004

This theorem depends on definitions:  df-bi 110  df-tru 1231  df-fal 1234  df-nf 1330  df-sb 1628  df-clab 2009  df-cleq 2015  df-clel 2018  df-in 2901  df-ss 2908

This theorem is referenced by: (None)