Theorem sspsstr 4057

Description: Transitive law for subclass and proper subclass. (Contributed by NM, 3-Apr-1996.)

Assertion

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

Proof of Theorem sspsstr

Step	Hyp	Ref	Expression
1		sspss 4050	. 2 ⊢ (𝐴 ⊆ 𝐵 ↔ (𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵))
2		psstr 4056	. . . . 5 ⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)
3	2	ex 418	. . . 4 ⊢ (𝐴 ⊊ 𝐵 → (𝐵 ⊊ 𝐶 → 𝐴 ⊊ 𝐶))
4		psseq1 4038	. . . . 5 ⊢ (𝐴 = 𝐵 → (𝐴 ⊊ 𝐶 ↔ 𝐵 ⊊ 𝐶))
5	4	biimprd 251	. . . 4 ⊢ (𝐴 = 𝐵 → (𝐵 ⊊ 𝐶 → 𝐴 ⊊ 𝐶))
6	3, 5	jaoi 871	. . 3 ⊢ ((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) → (𝐵 ⊊ 𝐶 → 𝐴 ⊊ 𝐶))
7	6	imp 412	. 2 ⊢ (((𝐴 ⊊ 𝐵 ∨ 𝐴 = 𝐵) ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)
8	1, 7	sylanb 593	1 ⊢ ((𝐴 ⊆ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∨ wo 861   = wceq 1570   ⊆ wss 3899   ⊊ wpss 3900

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-cleq 2753  df-ne 2957  df-ss 3916  df-pss 3919

This theorem is used by:  sspsstrd  4060  ordtr2  6401  php  9206  canthp1lem2  10719  suplem1pr  11118  fbfinnfr  24140  ppiltx  27486