Step |
Hyp |
Ref |
Expression |
1 |
|
eqcom |
⊢ ( 𝑦 = ( 𝐹 ‘ 𝑥 ) ↔ ( 𝐹 ‘ 𝑥 ) = 𝑦 ) |
2 |
|
tz6.12i |
⊢ ( 𝑦 ≠ ∅ → ( ( 𝐹 ‘ 𝑥 ) = 𝑦 → 𝑥 𝐹 𝑦 ) ) |
3 |
1 2
|
syl5bi |
⊢ ( 𝑦 ≠ ∅ → ( 𝑦 = ( 𝐹 ‘ 𝑥 ) → 𝑥 𝐹 𝑦 ) ) |
4 |
3
|
eximdv |
⊢ ( 𝑦 ≠ ∅ → ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → ∃ 𝑥 𝑥 𝐹 𝑦 ) ) |
5 |
|
vex |
⊢ 𝑦 ∈ V |
6 |
5
|
elrn |
⊢ ( 𝑦 ∈ ran 𝐹 ↔ ∃ 𝑥 𝑥 𝐹 𝑦 ) |
7 |
4 6
|
syl6ibr |
⊢ ( 𝑦 ≠ ∅ → ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → 𝑦 ∈ ran 𝐹 ) ) |
8 |
7
|
com12 |
⊢ ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → ( 𝑦 ≠ ∅ → 𝑦 ∈ ran 𝐹 ) ) |
9 |
8
|
necon1bd |
⊢ ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → ( ¬ 𝑦 ∈ ran 𝐹 → 𝑦 = ∅ ) ) |
10 |
|
velsn |
⊢ ( 𝑦 ∈ { ∅ } ↔ 𝑦 = ∅ ) |
11 |
9 10
|
syl6ibr |
⊢ ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → ( ¬ 𝑦 ∈ ran 𝐹 → 𝑦 ∈ { ∅ } ) ) |
12 |
11
|
orrd |
⊢ ( ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) → ( 𝑦 ∈ ran 𝐹 ∨ 𝑦 ∈ { ∅ } ) ) |
13 |
12
|
ss2abi |
⊢ { 𝑦 ∣ ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) } ⊆ { 𝑦 ∣ ( 𝑦 ∈ ran 𝐹 ∨ 𝑦 ∈ { ∅ } ) } |
14 |
|
df-un |
⊢ ( ran 𝐹 ∪ { ∅ } ) = { 𝑦 ∣ ( 𝑦 ∈ ran 𝐹 ∨ 𝑦 ∈ { ∅ } ) } |
15 |
13 14
|
sseqtrri |
⊢ { 𝑦 ∣ ∃ 𝑥 𝑦 = ( 𝐹 ‘ 𝑥 ) } ⊆ ( ran 𝐹 ∪ { ∅ } ) |