Theorem ssnelpss 1751

Description: A subclass missing a member is a proper subclass.

Assertion

Ref	Expression
ssnelpss	⊢ (A ⊆ B → ((C ∈ B ∧ ¬ C ∈ A) → A ⊂ B))

Proof of Theorem ssnelpss

Step	Hyp	Ref	Expression
1		dfpss2 1557	. . 3 ⊢ (A ⊂ B ↔ (A ⊆ B ∧ ¬ A = B))
2	1	baibr 507	. 2 ⊢ (A ⊆ B → (¬ A = B ↔ A ⊂ B))
3		clneq2 1169	. . 3 ⊢ ((C ∈ B ∧ ¬ C ∈ A) → ¬ B = A)
4		cleqcom 1103	. . . 4 ⊢ (B = A ↔ A = B)
5	4	negbii 162	. . 3 ⊢ (¬ B = A ↔ ¬ A = B)
6	3, 5	sylib 173	. 2 ⊢ ((C ∈ B ∧ ¬ C ∈ A) → ¬ A = B)
7	2, 6	syl5bi 183	1 ⊢ (A ⊆ B → ((C ∈ B ∧ ¬ C ∈ A) → A ⊂ B))

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ⊆ wss 1487   ⊂ wpss 1488

This theorem is referenced by:  nthruc 4784  nthruz 4785

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-gen 677  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-cleq 1097  df-clel 1099  df-ne 1192  df-pss 1494

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