Step |
Hyp |
Ref |
Expression |
1 |
|
ramval.c |
⊢ 𝐶 = ( 𝑎 ∈ V , 𝑖 ∈ ℕ0 ↦ { 𝑏 ∈ 𝒫 𝑎 ∣ ( ♯ ‘ 𝑏 ) = 𝑖 } ) |
2 |
|
0nn0 |
⊢ 0 ∈ ℕ0 |
3 |
1
|
hashbcval |
⊢ ( ( 𝐴 ∈ 𝑉 ∧ 0 ∈ ℕ0 ) → ( 𝐴 𝐶 0 ) = { 𝑥 ∈ 𝒫 𝐴 ∣ ( ♯ ‘ 𝑥 ) = 0 } ) |
4 |
2 3
|
mpan2 |
⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 𝐶 0 ) = { 𝑥 ∈ 𝒫 𝐴 ∣ ( ♯ ‘ 𝑥 ) = 0 } ) |
5 |
|
hasheq0 |
⊢ ( 𝑥 ∈ V → ( ( ♯ ‘ 𝑥 ) = 0 ↔ 𝑥 = ∅ ) ) |
6 |
5
|
elv |
⊢ ( ( ♯ ‘ 𝑥 ) = 0 ↔ 𝑥 = ∅ ) |
7 |
6
|
anbi2i |
⊢ ( ( 𝑥 ∈ 𝒫 𝐴 ∧ ( ♯ ‘ 𝑥 ) = 0 ) ↔ ( 𝑥 ∈ 𝒫 𝐴 ∧ 𝑥 = ∅ ) ) |
8 |
|
id |
⊢ ( 𝑥 = ∅ → 𝑥 = ∅ ) |
9 |
|
0elpw |
⊢ ∅ ∈ 𝒫 𝐴 |
10 |
8 9
|
eqeltrdi |
⊢ ( 𝑥 = ∅ → 𝑥 ∈ 𝒫 𝐴 ) |
11 |
10
|
pm4.71ri |
⊢ ( 𝑥 = ∅ ↔ ( 𝑥 ∈ 𝒫 𝐴 ∧ 𝑥 = ∅ ) ) |
12 |
7 11
|
bitr4i |
⊢ ( ( 𝑥 ∈ 𝒫 𝐴 ∧ ( ♯ ‘ 𝑥 ) = 0 ) ↔ 𝑥 = ∅ ) |
13 |
12
|
abbii |
⊢ { 𝑥 ∣ ( 𝑥 ∈ 𝒫 𝐴 ∧ ( ♯ ‘ 𝑥 ) = 0 ) } = { 𝑥 ∣ 𝑥 = ∅ } |
14 |
|
df-rab |
⊢ { 𝑥 ∈ 𝒫 𝐴 ∣ ( ♯ ‘ 𝑥 ) = 0 } = { 𝑥 ∣ ( 𝑥 ∈ 𝒫 𝐴 ∧ ( ♯ ‘ 𝑥 ) = 0 ) } |
15 |
|
df-sn |
⊢ { ∅ } = { 𝑥 ∣ 𝑥 = ∅ } |
16 |
13 14 15
|
3eqtr4i |
⊢ { 𝑥 ∈ 𝒫 𝐴 ∣ ( ♯ ‘ 𝑥 ) = 0 } = { ∅ } |
17 |
4 16
|
eqtrdi |
⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 𝐶 0 ) = { ∅ } ) |