HomeHome Metamath Proof Explorer < Previous   Next >
Related theorems
GIF version

Theorem xpundir 2462
Description: Distributive law for cross product over union. Similar to Theorem 103 of [Suppes] p. 52.
Assertion
Ref Expression
xpundir ((AB) × C) = ((A × C) ∪ (B × C))

Proof of Theorem xpundir
StepHypRef Expression
1 elun 1601 . . . . . 6 (x ∈ (AB) ↔ (xAxB))
21anbi1i 368 . . . . 5 ((x ∈ (AB) ∧ yC) ↔ ((xAxB) ∧ yC))
3 andir 457 . . . . 5 (((xAxB) ∧ yC) ↔ ((xAyC) ∨ (xByC)))
42, 3bitr 151 . . . 4 ((x ∈ (AB) ∧ yC) ↔ ((xAyC) ∨ (xByC)))
54biopabi 2103 . . 3 {⟨x, y⟩∣(x ∈ (AB) ∧ yC)} = {⟨x, y⟩∣((xAyC) ∨ (xByC))}
6 unopab 2121 . . 3 ({⟨x, y⟩∣(xAyC)} ∪ {⟨x, y⟩∣(xByC)}) = {⟨x, y⟩∣((xAyC) ∨ (xByC))}
75, 6eqtr4 1122 . 2 {⟨x, y⟩∣(x ∈ (AB) ∧ yC)} = ({⟨x, y⟩∣(xAyC)} ∪ {⟨x, y⟩∣(xByC)})
8 df-xp 2424 . 2 ((AB) × C) = {⟨x, y⟩∣(x ∈ (AB) ∧ yC)}
9 df-xp 2424 . . 3 (A × C) = {⟨x, y⟩∣(xAyC)}
10 df-xp 2424 . . 3 (B × C) = {⟨x, y⟩∣(xByC)}
119, 10uneq12i 1609 . 2 ((A × C) ∪ (B × C)) = ({⟨x, y⟩∣(xAyC)} ∪ {⟨x, y⟩∣(xByC)})
127, 8, 113eqtr4 1126 1 ((AB) × C) = ((A × C) ∪ (B × C))
Colors of variables: wff set class
Syntax hints:   ∨ wo 195   ∧ wa 196   = wceq 1091   ∈ wcel 1092   ∪ cun 1485  {copab 2055   × cxp 2408
This theorem is referenced by:  xpun 2463  resundi 2582  cdaassen 3725
This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074
This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-un 1490  df-opab 2098  df-xp 2424
metamath.org