Metamath Proof Explorer


Theorem disjpreima

Description: A preimage of a disjoint set is disjoint. (Contributed by Thierry Arnoux, 7-Feb-2017)

Ref Expression
Assertion disjpreima Fun F Disj x A B Disj x A F -1 B

Proof

Step Hyp Ref Expression
1 inpreima Fun F F -1 y / x B z / x B = F -1 y / x B F -1 z / x B
2 imaeq2 y / x B z / x B = F -1 y / x B z / x B = F -1
3 ima0 F -1 =
4 2 3 eqtrdi y / x B z / x B = F -1 y / x B z / x B =
5 1 4 sylan9req Fun F y / x B z / x B = F -1 y / x B F -1 z / x B =
6 5 ex Fun F y / x B z / x B = F -1 y / x B F -1 z / x B =
7 csbima12 y / x F -1 B = y / x F -1 y / x B
8 csbconstg y V y / x F -1 = F -1
9 8 elv y / x F -1 = F -1
10 9 imaeq1i y / x F -1 y / x B = F -1 y / x B
11 7 10 eqtri y / x F -1 B = F -1 y / x B
12 csbima12 z / x F -1 B = z / x F -1 z / x B
13 csbconstg z V z / x F -1 = F -1
14 13 elv z / x F -1 = F -1
15 14 imaeq1i z / x F -1 z / x B = F -1 z / x B
16 12 15 eqtri z / x F -1 B = F -1 z / x B
17 11 16 ineq12i y / x F -1 B z / x F -1 B = F -1 y / x B F -1 z / x B
18 17 eqeq1i y / x F -1 B z / x F -1 B = F -1 y / x B F -1 z / x B =
19 6 18 syl6ibr Fun F y / x B z / x B = y / x F -1 B z / x F -1 B =
20 19 orim2d Fun F y = z y / x B z / x B = y = z y / x F -1 B z / x F -1 B =
21 20 ralimdv Fun F z A y = z y / x B z / x B = z A y = z y / x F -1 B z / x F -1 B =
22 21 ralimdv Fun F y A z A y = z y / x B z / x B = y A z A y = z y / x F -1 B z / x F -1 B =
23 disjors Disj x A B y A z A y = z y / x B z / x B =
24 disjors Disj x A F -1 B y A z A y = z y / x F -1 B z / x F -1 B =
25 22 23 24 3imtr4g Fun F Disj x A B Disj x A F -1 B
26 25 imp Fun F Disj x A B Disj x A F -1 B