Metamath Proof Explorer


Theorem disji2f

Description: Property of a disjoint collection: if B ( x ) = C and B ( Y ) = D , and x =/= Y , then B and C are disjoint. (Contributed by Thierry Arnoux, 30-Dec-2016)

Ref Expression
Hypotheses disjif.1 _ x C
disjif.2 x = Y B = C
Assertion disji2f Disj x A B x A Y A x Y B C =

Proof

Step Hyp Ref Expression
1 disjif.1 _ x C
2 disjif.2 x = Y B = C
3 df-ne x Y ¬ x = Y
4 disjors Disj x A B y A z A y = z y / x B z / x B =
5 equequ1 y = x y = z x = z
6 csbeq1 y = x y / x B = x / x B
7 csbid x / x B = B
8 6 7 eqtrdi y = x y / x B = B
9 8 ineq1d y = x y / x B z / x B = B z / x B
10 9 eqeq1d y = x y / x B z / x B = B z / x B =
11 5 10 orbi12d y = x y = z y / x B z / x B = x = z B z / x B =
12 eqeq2 z = Y x = z x = Y
13 nfcv _ x Y
14 13 1 2 csbhypf z = Y z / x B = C
15 14 ineq2d z = Y B z / x B = B C
16 15 eqeq1d z = Y B z / x B = B C =
17 12 16 orbi12d z = Y x = z B z / x B = x = Y B C =
18 11 17 rspc2v x A Y A y A z A y = z y / x B z / x B = x = Y B C =
19 4 18 syl5bi x A Y A Disj x A B x = Y B C =
20 19 impcom Disj x A B x A Y A x = Y B C =
21 20 ord Disj x A B x A Y A ¬ x = Y B C =
22 3 21 syl5bi Disj x A B x A Y A x Y B C =
23 22 3impia Disj x A B x A Y A x Y B C =