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Theorem invdisj 3750
 Description: If there is a function 𝐶(y) such that 𝐶(y) = x for all y ∈ B(x), then the sets B(x) for distinct x ∈ A are disjoint. (Contributed by Mario Carneiro, 10-Dec-2016.)
Assertion
Ref Expression
invdisj (x A y B 𝐶 = xDisj x A B)
Distinct variable groups:   x,y   y,A   y,B   x,𝐶
Allowed substitution hints:   A(x)   B(x)   𝐶(y)

Proof of Theorem invdisj
StepHypRef Expression
1 nfra2xy 2358 . . 3 yx A y B 𝐶 = x
2 df-ral 2305 . . . . 5 (x A y B 𝐶 = xx(x Ay B 𝐶 = x))
3 rsp 2363 . . . . . . . . 9 (y B 𝐶 = x → (y B𝐶 = x))
4 eqcom 2039 . . . . . . . . 9 (𝐶 = xx = 𝐶)
53, 4syl6ib 150 . . . . . . . 8 (y B 𝐶 = x → (y Bx = 𝐶))
65imim2i 12 . . . . . . 7 ((x Ay B 𝐶 = x) → (x A → (y Bx = 𝐶)))
76impd 242 . . . . . 6 ((x Ay B 𝐶 = x) → ((x A y B) → x = 𝐶))
87alimi 1341 . . . . 5 (x(x Ay B 𝐶 = x) → x((x A y B) → x = 𝐶))
92, 8sylbi 114 . . . 4 (x A y B 𝐶 = xx((x A y B) → x = 𝐶))
10 mo2icl 2714 . . . 4 (x((x A y B) → x = 𝐶) → ∃*x(x A y B))
119, 10syl 14 . . 3 (x A y B 𝐶 = x∃*x(x A y B))
121, 11alrimi 1412 . 2 (x A y B 𝐶 = xy∃*x(x A y B))
13 dfdisj2 3738 . 2 (Disj x A By∃*x(x A y B))
1412, 13sylibr 137 1 (x A y B 𝐶 = xDisj x A B)
 Colors of variables: wff set class Syntax hints:   → wi 4   ∧ wa 97  ∀wal 1240   = wceq 1242   ∈ wcel 1390  ∃*wmo 1898  ∀wral 2300  Disj wdisj 3736 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019 This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-eu 1900  df-mo 1901  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-ral 2305  df-rmo 2308  df-v 2553  df-disj 3737 This theorem is referenced by: (None)
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