Metamath Proof Explorer


Theorem xpdjuen

Description: Cardinal multiplication distributes over cardinal addition. Theorem 6I(3) of Enderton p. 142. (Contributed by NM, 26-Sep-2004) (Revised by Mario Carneiro, 29-Apr-2015)

Ref Expression
Assertion xpdjuen A V B W C X A × B ⊔︀ C A × B ⊔︀ A × C

Proof

Step Hyp Ref Expression
1 enrefg A V A A
2 1 3ad2ant1 A V B W C X A A
3 0ex V
4 simp2 A V B W C X B W
5 xpsnen2g V B W × B B
6 3 4 5 sylancr A V B W C X × B B
7 6 ensymd A V B W C X B × B
8 xpen A A B × B A × B A × × B
9 2 7 8 syl2anc A V B W C X A × B A × × B
10 1on 1 𝑜 On
11 simp3 A V B W C X C X
12 xpsnen2g 1 𝑜 On C X 1 𝑜 × C C
13 10 11 12 sylancr A V B W C X 1 𝑜 × C C
14 13 ensymd A V B W C X C 1 𝑜 × C
15 xpen A A C 1 𝑜 × C A × C A × 1 𝑜 × C
16 2 14 15 syl2anc A V B W C X A × C A × 1 𝑜 × C
17 xp01disjl × B 1 𝑜 × C =
18 17 xpeq2i A × × B 1 𝑜 × C = A ×
19 xpindi A × × B 1 𝑜 × C = A × × B A × 1 𝑜 × C
20 xp0 A × =
21 18 19 20 3eqtr3i A × × B A × 1 𝑜 × C =
22 21 a1i A V B W C X A × × B A × 1 𝑜 × C =
23 djuenun A × B A × × B A × C A × 1 𝑜 × C A × × B A × 1 𝑜 × C = A × B ⊔︀ A × C A × × B A × 1 𝑜 × C
24 9 16 22 23 syl3anc A V B W C X A × B ⊔︀ A × C A × × B A × 1 𝑜 × C
25 df-dju B ⊔︀ C = × B 1 𝑜 × C
26 25 xpeq2i A × B ⊔︀ C = A × × B 1 𝑜 × C
27 xpundi A × × B 1 𝑜 × C = A × × B A × 1 𝑜 × C
28 26 27 eqtri A × B ⊔︀ C = A × × B A × 1 𝑜 × C
29 24 28 breqtrrdi A V B W C X A × B ⊔︀ A × C A × B ⊔︀ C
30 29 ensymd A V B W C X A × B ⊔︀ C A × B ⊔︀ A × C