Metamath Proof Explorer


Theorem ex-pw

Description: Example for df-pw . Example by David A. Wheeler. (Contributed by Mario Carneiro, 2-Jul-2016)

Ref Expression
Assertion ex-pw ⊢ A = 3 5 7 → 𝒫 A = ∅ ∪ 3 5 7 ∪ 3 5 3 7 5 7 ∪ 3 5 7

Proof

Step Hyp Ref Expression
1 pweq ⊢ A = 3 5 7 → 𝒫 A = 𝒫 3 5 7
2 qdass ⊢ ∅ 3 ∪ 5 3 5 = ∅ 3 5 ∪ 3 5
3 qdassr ⊢ 7 3 7 ∪ 5 7 3 5 7 = 7 ∪ 3 7 5 7 3 5 7
4 2 3 uneq12i ⊢ ∅ 3 ∪ 5 3 5 ∪ 7 3 7 ∪ 5 7 3 5 7 = ∅ 3 5 ∪ 3 5 ∪ 7 ∪ 3 7 5 7 3 5 7
5 pwtp ⊢ 𝒫 3 5 7 = ∅ 3 ∪ 5 3 5 ∪ 7 3 7 ∪ 5 7 3 5 7
6 df-tp ⊢ 3 5 7 = 3 5 ∪ 7
7 6 uneq2i ⊢ ∅ ∪ 3 5 7 = ∅ ∪ 3 5 ∪ 7
8 unass ⊢ ∅ ∪ 3 5 ∪ 7 = ∅ ∪ 3 5 ∪ 7
9 7 8 eqtr4i ⊢ ∅ ∪ 3 5 7 = ∅ ∪ 3 5 ∪ 7
10 tpass ⊢ ∅ 3 5 = ∅ ∪ 3 5
11 10 uneq1i ⊢ ∅ 3 5 ∪ 7 = ∅ ∪ 3 5 ∪ 7
12 9 11 eqtr4i ⊢ ∅ ∪ 3 5 7 = ∅ 3 5 ∪ 7
13 unass ⊢ 3 5 ∪ 3 7 5 7 ∪ 3 5 7 = 3 5 ∪ 3 7 5 7 ∪ 3 5 7
14 tpass ⊢ 3 5 3 7 5 7 = 3 5 ∪ 3 7 5 7
15 14 uneq1i ⊢ 3 5 3 7 5 7 ∪ 3 5 7 = 3 5 ∪ 3 7 5 7 ∪ 3 5 7
16 df-tp ⊢ 3 7 5 7 3 5 7 = 3 7 5 7 ∪ 3 5 7
17 16 uneq2i ⊢ 3 5 ∪ 3 7 5 7 3 5 7 = 3 5 ∪ 3 7 5 7 ∪ 3 5 7
18 13 15 17 3eqtr4i ⊢ 3 5 3 7 5 7 ∪ 3 5 7 = 3 5 ∪ 3 7 5 7 3 5 7
19 12 18 uneq12i ⊢ ∅ ∪ 3 5 7 ∪ 3 5 3 7 5 7 ∪ 3 5 7 = ∅ 3 5 ∪ 7 ∪ 3 5 ∪ 3 7 5 7 3 5 7
20 un4 ⊢ ∅ 3 5 ∪ 3 5 ∪ 7 ∪ 3 7 5 7 3 5 7 = ∅ 3 5 ∪ 7 ∪ 3 5 ∪ 3 7 5 7 3 5 7
21 19 20 eqtr4i ⊢ ∅ ∪ 3 5 7 ∪ 3 5 3 7 5 7 ∪ 3 5 7 = ∅ 3 5 ∪ 3 5 ∪ 7 ∪ 3 7 5 7 3 5 7
22 4 5 21 3eqtr4i ⊢ 𝒫 3 5 7 = ∅ ∪ 3 5 7 ∪ 3 5 3 7 5 7 ∪ 3 5 7
23 1 22 eqtrdi ⊢ A = 3 5 7 → 𝒫 A = ∅ ∪ 3 5 7 ∪ 3 5 3 7 5 7 ∪ 3 5 7