Metamath Proof Explorer


Theorem enp1i

Description: Proof induction for en2 and related theorems. (Contributed by Mario Carneiro, 5-Jan-2016) Generalize to all ordinals and avoid ax-pow , ax-un . (Revised by BTernaryTau, 6-Jan-2025)

Ref Expression
Hypotheses enp1i.1 ⊢ Ord ⁡ M
enp1i.2 ⊢ N = suc ⁡ M
enp1i.3 ⊢ A ∖ x ≈ M → φ
enp1i.4 ⊢ x ∈ A → φ → ψ
Assertion enp1i ⊢ A ≈ N → ∃ x ψ

Proof

Step Hyp Ref Expression
1 enp1i.1 ⊢ Ord ⁡ M
2 enp1i.2 ⊢ N = suc ⁡ M
3 enp1i.3 ⊢ A ∖ x ≈ M → φ
4 enp1i.4 ⊢ x ∈ A → φ → ψ
5 2 breq2i ⊢ A ≈ N ↔ A ≈ suc ⁡ M
6 encv ⊢ A ≈ suc ⁡ M → A ∈ V ∧ suc ⁡ M ∈ V
7 6 simprd ⊢ A ≈ suc ⁡ M → suc ⁡ M ∈ V
8 sssucid ⊢ M ⊆ suc ⁡ M
9 ssexg ⊢ M ⊆ suc ⁡ M ∧ suc ⁡ M ∈ V → M ∈ V
10 8 9 mpan ⊢ suc ⁡ M ∈ V → M ∈ V
11 elong ⊢ M ∈ V → M ∈ On ↔ Ord ⁡ M
12 7 10 11 3syl ⊢ A ≈ suc ⁡ M → M ∈ On ↔ Ord ⁡ M
13 1 12 mpbiri ⊢ A ≈ suc ⁡ M → M ∈ On
14 rexdif1en ⊢ M ∈ On ∧ A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
15 13 14 mpancom ⊢ A ≈ suc ⁡ M → ∃ x ∈ A A ∖ x ≈ M
16 3 reximi ⊢ ∃ x ∈ A A ∖ x ≈ M → ∃ x ∈ A φ
17 df-rex ⊢ ∃ x ∈ A φ ↔ ∃ x x ∈ A ∧ φ
18 4 imp ⊢ x ∈ A ∧ φ → ψ
19 18 eximi ⊢ ∃ x x ∈ A ∧ φ → ∃ x ψ
20 17 19 sylbi ⊢ ∃ x ∈ A φ → ∃ x ψ
21 15 16 20 3syl ⊢ A ≈ suc ⁡ M → ∃ x ψ
22 5 21 sylbi ⊢ A ≈ N → ∃ x ψ