Metamath Proof Explorer


Theorem bnj106

Description: First-order logic and set theory. (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj106.1 ⊢ ψ ↔ ∀ i ∈ ω suc ⁡ i ∈ n → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R
bnj106.2 ⊢ F ∈ V
Assertion bnj106 ⊢ [˙F / f]˙ [˙ 1 𝑜 / n]˙ ψ ↔ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R

Proof

Step Hyp Ref Expression
1 bnj106.1 ⊢ ψ ↔ ∀ i ∈ ω suc ⁡ i ∈ n → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R
2 bnj106.2 ⊢ F ∈ V
3 bnj105 ⊢ 1 𝑜 ∈ V
4 1 3 bnj92 ⊢ [˙ 1 𝑜 / n]˙ ψ ↔ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R
5 4 sbcbii ⊢ [˙F / f]˙ [˙ 1 𝑜 / n]˙ ψ ↔ [˙F / f]˙ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R
6 fveq1 ⊢ f = F → f ⁡ suc ⁡ i = F ⁡ suc ⁡ i
7 fveq1 ⊢ f = F → f ⁡ i = F ⁡ i
8 7 bnj1113 ⊢ f = F → ⋃ y ∈ f ⁡ i pred y A R = ⋃ y ∈ F ⁡ i pred y A R
9 6 8 eqeq12d ⊢ f = F → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R ↔ F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R
10 9 imbi2d ⊢ f = F → suc ⁡ i ∈ 1 𝑜 → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R ↔ suc ⁡ i ∈ 1 𝑜 → F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R
11 10 ralbidv ⊢ f = F → ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R ↔ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R
12 2 11 sbcie ⊢ [˙F / f]˙ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → f ⁡ suc ⁡ i = ⋃ y ∈ f ⁡ i pred y A R ↔ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R
13 5 12 bitri ⊢ [˙F / f]˙ [˙ 1 𝑜 / n]˙ ψ ↔ ∀ i ∈ ω suc ⁡ i ∈ 1 𝑜 → F ⁡ suc ⁡ i = ⋃ y ∈ F ⁡ i pred y A R