Theorem npss 4136

Description: A class is not a proper subclass of another iff it satisfies a one-directional form of eqss 4024. (Contributed by Mario Carneiro, 15-May-2015.)

Assertion

Ref	Expression
npss	⊢ (¬ 𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 → 𝐴 = 𝐵))

Proof of Theorem npss

Step	Hyp	Ref	Expression
1		pm4.61 404	. . 3 ⊢ (¬ (𝐴 ⊆ 𝐵 → 𝐴 = 𝐵) ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
2		dfpss2 4111	. . 3 ⊢ (𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 ∧ ¬ 𝐴 = 𝐵))
3	1, 2	bitr4i 278	. 2 ⊢ (¬ (𝐴 ⊆ 𝐵 → 𝐴 = 𝐵) ↔ 𝐴 ⊊ 𝐵)
4	3	con1bii 356	1 ⊢ (¬ 𝐴 ⊊ 𝐵 ↔ (𝐴 ⊆ 𝐵 → 𝐴 = 𝐵))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206   ∧ wa 395   = wceq 1537   ⊆ wss 3976   ⊊ wpss 3977

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207  df-an 396  df-ne 2947  df-pss 3996

This theorem is referenced by:  ttukeylem7  10584  canthp1lem2  10722  pgpfac1lem1  20118  lspsncv0  21171  obslbs  21773  ssmxidl  33467  fvineqsneq  37378