Theorem nssr 3207

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 1640	. 2 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
2		dfss2 3136	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 → 𝑥 ∈ 𝐵))
3	1, 2	sylnibr 672	1 ⊢ (∃𝑥(𝑥 ∈ 𝐴 ∧ ¬ 𝑥 ∈ 𝐵) → ¬ 𝐴 ⊆ 𝐵)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103  ∀wal 1346  ∃wex 1485   ∈ wcel 2141   ⊆ wss 3121

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 609  ax-in2 610  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-11 1499  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-fal 1354  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-in 3127  df-ss 3134

This theorem is referenced by: (None)