Theorem psstr 4065

Description: Transitive law for proper subclass. Theorem 9 of [Suppes] p. 23. (Contributed by NM, 7-Feb-1996.)

Assertion

Ref	Expression
psstr	⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)

Proof of Theorem psstr

Step	Hyp	Ref	Expression
1		pssss 4055	. . 3 ⊢ (𝐴 ⊊ 𝐵 → 𝐴 ⊆ 𝐵)
2		pssss 4055	. . 3 ⊢ (𝐵 ⊊ 𝐶 → 𝐵 ⊆ 𝐶)
3	1, 2	sylan9ss 3953	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊆ 𝐶)
4		pssn2lp 4062	. . . 4 ⊢ ¬ (𝐶 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶)
5		psseq1 4047	. . . . 5 ⊢ (𝐴 = 𝐶 → (𝐴 ⊊ 𝐵 ↔ 𝐶 ⊊ 𝐵))
6	5	anbi1d 643	. . . 4 ⊢ (𝐴 = 𝐶 → ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) ↔ (𝐶 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶)))
7	4, 6	mtbiri 330	. . 3 ⊢ (𝐴 = 𝐶 → ¬ (𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶))
8	7	con2i 140	. 2 ⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → ¬ 𝐴 = 𝐶)
9		dfpss2 4045	. 2 ⊢ (𝐴 ⊊ 𝐶 ↔ (𝐴 ⊆ 𝐶 ∧ ¬ 𝐴 = 𝐶))
10	3, 8, 9	sylanbrc 595	1 ⊢ ((𝐴 ⊊ 𝐵 ∧ 𝐵 ⊊ 𝐶) → 𝐴 ⊊ 𝐶)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ⊆ wss 3908   ⊊ wpss 3909

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 2156  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-cleq 2758  df-ne 2962  df-ss 3925  df-pss 3928

This theorem is used by:  sspsstr  4066  psssstr  4067  psstrd  4068  porpss  7737  inf3lem5  9611  ltsopr  11035