Theorem sspsstr 1575

Description: Transitive law for subclass and proper subclass.

Assertion

Ref	Expression
sspsstr	⊢ ((A ⊆ B ∧ B ⊂ C) → A ⊂ C)

Proof of Theorem sspsstr

Step	Hyp	Ref	Expression
1		psstr 1574	. . . . 5 ⊢ ((A ⊂ B ∧ B ⊂ C) → A ⊂ C)
2	1	exp 291	. . . 4 ⊢ (A ⊂ B → (B ⊂ C → A ⊂ C))
3		psseq1 1559	. . . . 5 ⊢ (A = B → (A ⊂ C ↔ B ⊂ C))
4	3	biimprd 136	. . . 4 ⊢ (A = B → (B ⊂ C → A ⊂ C))
5	2, 4	jaoi 275	. . 3 ⊢ ((A ⊂ B ∨ A = B) → (B ⊂ C → A ⊂ C))
6	5	imp 277	. 2 ⊢ (((A ⊂ B ∨ A = B) ∧ B ⊂ C) → A ⊂ C)
7		sspss 1569	. 2 ⊢ (A ⊆ B ↔ (A ⊂ B ∨ A = B))
8	6, 7	sylanb 344	1 ⊢ ((A ⊆ B ∧ B ⊂ C) → A ⊂ C)

Syntax hints:   → wi 2   ∨ wo 195   ∧ wa 196   = wceq 1091   ⊆ wss 1487   ⊂ wpss 1488

This theorem is referenced by:  php 3409  ltexprlem2 3937  suplem1pr 3955

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-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ne 1192  df-in 1491  df-ss 1492  df-pss 1494

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