Theorem nepss 36674

Description: Two classes are unequal iff their intersection is a proper subset of one of them. (Contributed by Scott Fenton, 23-Feb-2011.)

Assertion

Ref	Expression
nepss	⊢ (𝐴 ≠ 𝐵 ↔ ((𝐴 ∩ 𝐵) ⊊ 𝐴 ∨ (𝐴 ∩ 𝐵) ⊊ 𝐵))

Proof of Theorem nepss

Step	Hyp	Ref	Expression
1		nne 2963	. . . . . 6 ⊢ (¬ (𝐴 ∩ 𝐵) ≠ 𝐴 ↔ (𝐴 ∩ 𝐵) = 𝐴)
2		neeq1 3021	. . . . . . 7 ⊢ ((𝐴 ∩ 𝐵) = 𝐴 → ((𝐴 ∩ 𝐵) ≠ 𝐵 ↔ 𝐴 ≠ 𝐵))
3	2	biimprcd 253	. . . . . 6 ⊢ (𝐴 ≠ 𝐵 → ((𝐴 ∩ 𝐵) = 𝐴 → (𝐴 ∩ 𝐵) ≠ 𝐵))
4	1, 3	biimtrid 245	. . . . 5 ⊢ (𝐴 ≠ 𝐵 → (¬ (𝐴 ∩ 𝐵) ≠ 𝐴 → (𝐴 ∩ 𝐵) ≠ 𝐵))
5	4	orrd 877	. . . 4 ⊢ (𝐴 ≠ 𝐵 → ((𝐴 ∩ 𝐵) ≠ 𝐴 ∨ (𝐴 ∩ 𝐵) ≠ 𝐵))
6		inss1 4136	. . . . . 6 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐴
7	6	jctl 533	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ 𝐴 → ((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴))
8		inss2 4137	. . . . . 6 ⊢ (𝐴 ∩ 𝐵) ⊆ 𝐵
9	8	jctl 533	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ 𝐵 → ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵))
10	7, 9	orim12i 922	. . . 4 ⊢ (((𝐴 ∩ 𝐵) ≠ 𝐴 ∨ (𝐴 ∩ 𝐵) ≠ 𝐵) → (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) ∨ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵)))
11	5, 10	syl 18	. . 3 ⊢ (𝐴 ≠ 𝐵 → (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) ∨ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵)))
12		ineq2 4114	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐴) = (𝐴 ∩ 𝐵))
13		inidm 4126	. . . . . . 7 ⊢ (𝐴 ∩ 𝐴) = 𝐴
14	12, 13	eqtr3di 2813	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐴)
15	14	necon3i 2991	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ 𝐴 → 𝐴 ≠ 𝐵)
16	15	adantl 487	. . . 4 ⊢ (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) → 𝐴 ≠ 𝐵)
17		ineq1 4113	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = (𝐵 ∩ 𝐵))
18		inidm 4126	. . . . . . 7 ⊢ (𝐵 ∩ 𝐵) = 𝐵
19	17, 18	eqtrdi 2814	. . . . . 6 ⊢ (𝐴 = 𝐵 → (𝐴 ∩ 𝐵) = 𝐵)
20	19	necon3i 2991	. . . . 5 ⊢ ((𝐴 ∩ 𝐵) ≠ 𝐵 → 𝐴 ≠ 𝐵)
21	20	adantl 487	. . . 4 ⊢ (((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵) → 𝐴 ≠ 𝐵)
22	16, 21	jaoi 871	. . 3 ⊢ ((((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) ∨ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵)) → 𝐴 ≠ 𝐵)
23	11, 22	impbii 212	. 2 ⊢ (𝐴 ≠ 𝐵 ↔ (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) ∨ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵)))
24		df-pss 4017	. . 3 ⊢ ((𝐴 ∩ 𝐵) ⊊ 𝐴 ↔ ((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴))
25		df-pss 4017	. . 3 ⊢ ((𝐴 ∩ 𝐵) ⊊ 𝐵 ↔ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵))
26	24, 25	orbi12i 928	. 2 ⊢ (((𝐴 ∩ 𝐵) ⊊ 𝐴 ∨ (𝐴 ∩ 𝐵) ⊊ 𝐵) ↔ (((𝐴 ∩ 𝐵) ⊆ 𝐴 ∧ (𝐴 ∩ 𝐵) ≠ 𝐴) ∨ ((𝐴 ∩ 𝐵) ⊆ 𝐵 ∧ (𝐴 ∩ 𝐵) ≠ 𝐵)))
27	23, 26	bitr4i 281	1 ⊢ (𝐴 ≠ 𝐵 ↔ ((𝐴 ∩ 𝐵) ⊊ 𝐴 ∨ (𝐴 ∩ 𝐵) ⊊ 𝐵))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ≠ wne 2959   ⊆ wss 3900   ⊊ wpss 4016   ∩ cin 4092

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-rab 3417  df-v 3456  df-ss 3901  df-pss 4017  df-in 4094

This theorem is used by: (None)