Theorem disjneOLD 2920

Description: Members of disjoint sets are not equal.

Assertion

Ref	Expression
disjneOLD	|- (((A i^i B) = (/) /\ C e. A /\ D e. B) -> C =/= D)

Proof of Theorem disjneOLD

Step	Hyp	Ref	Expression
1		nelneq 1985	. . . . . . 7 |- ((D e. B /\ -. C e. B) -> -. D = C)
2		df-ne 2019	. . . . . . 7 |- (D =/= C <-> -. D = C)
3	1, 2	sylibr 217	. . . . . 6 |- ((D e. B /\ -. C e. B) -> D =/= C)
4	3	necomd 2095	. . . . 5 |- ((D e. B /\ -. C e. B) -> C =/= D)
5	4	ex 402	. . . 4 |- (D e. B -> (-. C e. B -> C =/= D))
6		eleq1 1957	. . . . . 6 |- (x = C -> (x e. B <-> C e. B))
7	6	notbid 673	. . . . 5 |- (x = C -> (-. x e. B <-> -. C e. B))
8	7	rcla4cva 2379	. . . 4 |- ((A.x e. A -. x e. B /\ C e. A) -> -. C e. B)
9	5, 8	syl5com 63	. . 3 |- ((A.x e. A -. x e. B /\ C e. A) -> (D e. B -> C =/= D))
10		disj 2914	. . 3 |- ((A i^i B) = (/) <-> A.x e. A -. x e. B)
11	9, 10	sylanb 498	. 2 |- (((A i^i B) = (/) /\ C e. A) -> (D e. B -> C =/= D))
12	11	3impia 1064	1 |- (((A i^i B) = (/) /\ C e. A /\ D e. B) -> C =/= D)

Syntax hints:  -. wn 2   -> wi 3   /\ wa 240   /\ w3a 858   = wceq 1298   e. wcel 1300   =/= wne 2017  A.wral 2105   i^i cin 2592  (/)c0 2875

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 1304  ax-gen 1305  ax-8 1306  ax-9 1307  ax-10 1308  ax-11 1309  ax-12 1310  ax-17 1317  ax-4 1319  ax-5o 1321  ax-6o 1324  ax-9o 1481  ax-10o 1500  ax-16 1580  ax-11o 1588  ax-ext 1865

This theorem depends on definitions:  df-bi 164  df-or 241  df-an 242  df-3an 860  df-ex 1327  df-sb 1536  df-clab 1872  df-cleq 1877  df-clel 1880  df-ne 2019  df-ral 2109  df-v 2294  df-dif 2597  df-in 2603  df-nul 2876

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