Metamath Proof Explorer


Theorem ex-xp

Description: Example for df-xp . Example by David A. Wheeler. (Contributed by Mario Carneiro, 7-May-2015)

Ref Expression
Assertion ex-xp ⊢ 1 5 × 2 7 = 1 2 1 7 ∪ 5 2 5 7

Proof

Step Hyp Ref Expression
1 df-pr ⊢ 1 5 = 1 ∪ 5
2 df-pr ⊢ 2 7 = 2 ∪ 7
3 1 2 xpeq12i ⊢ 1 5 × 2 7 = 1 ∪ 5 × 2 ∪ 7
4 xpun ⊢ 1 ∪ 5 × 2 ∪ 7 = 1 × 2 ∪ 1 × 7 ∪ 5 × 2 ∪ 5 × 7
5 1ex ⊢ 1 ∈ V
6 2nn ⊢ 2 ∈ ℕ
7 6 elexi ⊢ 2 ∈ V
8 5 7 xpsn ⊢ 1 × 2 = 1 2
9 7nn ⊢ 7 ∈ ℕ
10 9 elexi ⊢ 7 ∈ V
11 5 10 xpsn ⊢ 1 × 7 = 1 7
12 8 11 uneq12i ⊢ 1 × 2 ∪ 1 × 7 = 1 2 ∪ 1 7
13 df-pr ⊢ 1 2 1 7 = 1 2 ∪ 1 7
14 12 13 eqtr4i ⊢ 1 × 2 ∪ 1 × 7 = 1 2 1 7
15 5nn ⊢ 5 ∈ ℕ
16 15 elexi ⊢ 5 ∈ V
17 16 7 xpsn ⊢ 5 × 2 = 5 2
18 16 10 xpsn ⊢ 5 × 7 = 5 7
19 17 18 uneq12i ⊢ 5 × 2 ∪ 5 × 7 = 5 2 ∪ 5 7
20 df-pr ⊢ 5 2 5 7 = 5 2 ∪ 5 7
21 19 20 eqtr4i ⊢ 5 × 2 ∪ 5 × 7 = 5 2 5 7
22 14 21 uneq12i ⊢ 1 × 2 ∪ 1 × 7 ∪ 5 × 2 ∪ 5 × 7 = 1 2 1 7 ∪ 5 2 5 7
23 3 4 22 3eqtri ⊢ 1 5 × 2 7 = 1 2 1 7 ∪ 5 2 5 7