Metamath Proof Explorer


Theorem evl1fval1lem

Description: Lemma for evl1fval1 . (Contributed by AV, 11-Sep-2019)

Ref Expression
Hypotheses evl1fval1.q ⊢ Q = eval 1 ⁡ R
evl1fval1.b ⊢ B = Base R
Assertion evl1fval1lem ⊢ R ∈ V → Q = R evalSub 1 B

Proof

Step Hyp Ref Expression
1 evl1fval1.q ⊢ Q = eval 1 ⁡ R
2 evl1fval1.b ⊢ B = Base R
3 eqid ⊢ eval 1 ⁡ R = eval 1 ⁡ R
4 eqid ⊢ 1 𝑜 eval R = 1 𝑜 eval R
5 3 4 2 evl1fval ⊢ eval 1 ⁡ R = x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 eval R
6 1 a1i ⊢ R ∈ V → Q = eval 1 ⁡ R
7 2 fvexi ⊢ B ∈ V
8 7 pwid ⊢ B ∈ 𝒫 B
9 eqid ⊢ R evalSub 1 B = R evalSub 1 B
10 eqid ⊢ 1 𝑜 evalSub R = 1 𝑜 evalSub R
11 9 10 2 evls1fval ⊢ R ∈ V ∧ B ∈ 𝒫 B → R evalSub 1 B = x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 evalSub R ⁡ B
12 8 11 mpan2 ⊢ R ∈ V → R evalSub 1 B = x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 evalSub R ⁡ B
13 4 2 evlval ⊢ 1 𝑜 eval R = 1 𝑜 evalSub R ⁡ B
14 13 eqcomi ⊢ 1 𝑜 evalSub R ⁡ B = 1 𝑜 eval R
15 14 coeq2i ⊢ x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 evalSub R ⁡ B = x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 eval R
16 12 15 eqtrdi ⊢ R ∈ V → R evalSub 1 B = x ∈ B B 1 𝑜 ⟼ x ∘ y ∈ B ⟼ 1 𝑜 × y ∘ 1 𝑜 eval R
17 5 6 16 3eqtr4a ⊢ R ∈ V → Q = R evalSub 1 B