Theorem nssr 3162

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 1627	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		dfss2 3091	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 667	1 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ 𝐴 ⊆ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  ∀wal 1330  ∃wex 1469   ∈ wcel 1481   ⊆ wss 3076

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 604  ax-in2 605  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-11 1485  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516  ax-ext 2122

This theorem depends on definitions:  df-bi 116  df-tru 1335  df-fal 1338  df-nf 1438  df-sb 1737  df-clab 2127  df-cleq 2133  df-clel 2136  df-in 3082  df-ss 3089

This theorem is referenced by: (None)