Step |
Hyp |
Ref |
Expression |
1 |
|
peano2 |
⊢ ( 𝑧 ∈ ω → suc 𝑧 ∈ ω ) |
2 |
|
sseq1 |
⊢ ( 𝑥 = suc 𝑧 → ( 𝑥 ⊆ 𝑦 ↔ suc 𝑧 ⊆ 𝑦 ) ) |
3 |
2
|
rexbidv |
⊢ ( 𝑥 = suc 𝑧 → ( ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 ↔ ∃ 𝑦 ∈ 𝐴 suc 𝑧 ⊆ 𝑦 ) ) |
4 |
3
|
rspcv |
⊢ ( suc 𝑧 ∈ ω → ( ∀ 𝑥 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 → ∃ 𝑦 ∈ 𝐴 suc 𝑧 ⊆ 𝑦 ) ) |
5 |
|
sucssel |
⊢ ( 𝑧 ∈ V → ( suc 𝑧 ⊆ 𝑦 → 𝑧 ∈ 𝑦 ) ) |
6 |
5
|
elv |
⊢ ( suc 𝑧 ⊆ 𝑦 → 𝑧 ∈ 𝑦 ) |
7 |
6
|
reximi |
⊢ ( ∃ 𝑦 ∈ 𝐴 suc 𝑧 ⊆ 𝑦 → ∃ 𝑦 ∈ 𝐴 𝑧 ∈ 𝑦 ) |
8 |
4 7
|
syl6com |
⊢ ( ∀ 𝑥 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 → ( suc 𝑧 ∈ ω → ∃ 𝑦 ∈ 𝐴 𝑧 ∈ 𝑦 ) ) |
9 |
1 8
|
syl5 |
⊢ ( ∀ 𝑥 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 → ( 𝑧 ∈ ω → ∃ 𝑦 ∈ 𝐴 𝑧 ∈ 𝑦 ) ) |
10 |
9
|
ralrimiv |
⊢ ( ∀ 𝑥 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 → ∀ 𝑧 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑧 ∈ 𝑦 ) |
11 |
|
unbnn |
⊢ ( ( ω ∈ V ∧ 𝐴 ⊆ ω ∧ ∀ 𝑧 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑧 ∈ 𝑦 ) → 𝐴 ≈ ω ) |
12 |
10 11
|
syl3an3 |
⊢ ( ( ω ∈ V ∧ 𝐴 ⊆ ω ∧ ∀ 𝑥 ∈ ω ∃ 𝑦 ∈ 𝐴 𝑥 ⊆ 𝑦 ) → 𝐴 ≈ ω ) |