Theorem disjne 2315

Description: Members of disjoint sets are not equal.

Assertion

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

Proof of Theorem disjne

Step	Hyp	Ref	Expression
1		nelneq 1561	. . . . . . 7 |- ((D e. B /\ -. C e. B) -> -. D = C)
2		df-ne 1587	. . . . . . 7 |- (D =/= C <-> -. D = C)
3	1, 2	sylibr 200	. . . . . 6 |- ((D e. B /\ -. C e. B) -> D =/= C)
4	3	necomd 1637	. . . . 5 |- ((D e. B /\ -. C e. B) -> C =/= D)
5	4	ex 373	. . . 4 |- (D e. B -> (-. C e. B -> C =/= D))
6		eleq1 1534	. . . . . 6 |- (x = C -> (x e. B <-> C e. B))
7	6	negbid 611	. . . . 5 |- (x = C -> (-. x e. B <-> -. C e. B))
8	7	rcla4cva 1876	. . . 4 |- ((A.x e. A -. x e. B /\ C e. A) -> -. C e. B)
9	5, 8	syl5com 52	. . 3 |- ((A.x e. A -. x e. B /\ C e. A) -> (D e. B -> C =/= D))
10		disj 2311	. . 3 |- ((A i^i B) = (/) <-> A.x e. A -. x e. B)
11	9, 10	sylanb 449	. 2 |- (((A i^i B) = (/) /\ C e. A) -> (D e. B -> C =/= D))
12	11	3impia 830	1 |- (((A i^i B) = (/) /\ C e. A /\ D e. B) -> C =/= D)

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   /\ w3a 775   = wceq 956   e. wcel 958   =/= wne 1585  A.wral 1645   i^i cin 2046  (/)c0 2280

This theorem is referenced by:  brdom7disj 4804  brdom6disj 4805

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 777  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-ne 1587  df-ral 1649  df-v 1812  df-dif 2049  df-in 2051  df-nul 2281

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