Theorem disjne 2319

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 1564	. . . . . . 7 |- ((D e. B /\ -. C e. B) -> -. D = C)
2		df-ne 1590	. . . . . . 7 |- (D =/= C <-> -. D = C)
3	1, 2	sylibr 200	. . . . . 6 |- ((D e. B /\ -. C e. B) -> D =/= C)
4	3	necomd 1640	. . . . 5 |- ((D e. B /\ -. C e. B) -> C =/= D)
5	4	ex 373	. . . 4 |- (D e. B -> (-. C e. B -> C =/= D))
6		eleq1 1537	. . . . . 6 |- (x = C -> (x e. B <-> C e. B))
7	6	negbid 613	. . . . 5 |- (x = C -> (-. x e. B <-> -. C e. B))
8	7	rcla4cva 1879	. . . 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 2315	. . 3 |- ((A i^i B) = (/) <-> A.x e. A -. x e. B)
11	9, 10	sylanb 451	. 2 |- (((A i^i B) = (/) /\ C e. A) -> (D e. B -> C =/= D))
12	11	3impia 832	1 |- (((A i^i B) = (/) /\ C e. A /\ D e. B) -> C =/= D)

Syntax hints:  -. wn 2   -> wi 3   /\ wa 223   /\ w3a 777   = wceq 958   e. wcel 960   =/= wne 1588  A.wral 1648   i^i cin 2049  (/)c0 2283

This theorem is referenced by:  brdom7disj 4814  brdom6disj 4815

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 964  ax-gen 965  ax-8 966  ax-10 968  ax-12 970  ax-17 973  ax-4 975  ax-5o 977  ax-6o 980  ax-9o 1125  ax-10o 1142  ax-16 1212  ax-11o 1220  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-or 224  df-an 225  df-3an 779  df-ex 983  df-sb 1174  df-clab 1467  df-cleq 1472  df-clel 1475  df-ne 1590  df-ral 1652  df-v 1815  df-dif 2052  df-in 2054  df-nul 2284

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