Theorem psssstr 4058

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

Assertion

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

Proof of Theorem psssstr

Step	Hyp	Ref	Expression
1		sspss 4050	. 2 ⊢ (𝐵 ⊆ 𝐶 ↔ (𝐵 ⊊ 𝐶 ∨ 𝐵 = 𝐶))
2		psstr 4056	. . . . 5 ⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)
3	2	ex 418	. . . 4 ⊢ (𝐴 ⊊ 𝐵 → (𝐵 ⊊ 𝐶 → 𝐴 ⊊ 𝐶))
4		psseq2 4039	. . . . 5 ⊢ (𝐵 = 𝐶 → (𝐴 ⊊ 𝐵 ↔ 𝐴 ⊊ 𝐶))
5	4	biimpcd 252	. . . 4 ⊢ (𝐴 ⊊ 𝐵 → (𝐵 = 𝐶 → 𝐴 ⊊ 𝐶))
6	3, 5	jaod 873	. . 3 ⊢ (𝐴 ⊊ 𝐵 → ((𝐵 ⊊ 𝐶 ∨ 𝐵 = 𝐶) → 𝐴 ⊊ 𝐶))
7	6	imp 412	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∧ (𝐵 ⊊ 𝐶 ∨ 𝐵 = 𝐶)) → 𝐴 ⊊ 𝐶)
8	1, 7	sylan2b 606	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:  psssstrd  4061  suplem1pr  11137  atexch  32983  bj-2upln0  37936  bj-2upln1upl  37937