Metamath Proof Explorer


Theorem ruclem4

Description: Lemma for ruc . Initial value of the interval sequence. (Contributed by Mario Carneiro, 28-May-2014)

Ref Expression
Hypotheses ruc.1 ( 𝜑𝐹 : ℕ ⟶ ℝ )
ruc.2 ( 𝜑𝐷 = ( 𝑥 ∈ ( ℝ × ℝ ) , 𝑦 ∈ ℝ ↦ ( ( ( 1st𝑥 ) + ( 2nd𝑥 ) ) / 2 ) / 𝑚 if ( 𝑚 < 𝑦 , ⟨ ( 1st𝑥 ) , 𝑚 ⟩ , ⟨ ( ( 𝑚 + ( 2nd𝑥 ) ) / 2 ) , ( 2nd𝑥 ) ⟩ ) ) )
ruc.4 𝐶 = ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ 𝐹 )
ruc.5 𝐺 = seq 0 ( 𝐷 , 𝐶 )
Assertion ruclem4 ( 𝜑 → ( 𝐺 ‘ 0 ) = ⟨ 0 , 1 ⟩ )

Proof

Step Hyp Ref Expression
1 ruc.1 ( 𝜑𝐹 : ℕ ⟶ ℝ )
2 ruc.2 ( 𝜑𝐷 = ( 𝑥 ∈ ( ℝ × ℝ ) , 𝑦 ∈ ℝ ↦ ( ( ( 1st𝑥 ) + ( 2nd𝑥 ) ) / 2 ) / 𝑚 if ( 𝑚 < 𝑦 , ⟨ ( 1st𝑥 ) , 𝑚 ⟩ , ⟨ ( ( 𝑚 + ( 2nd𝑥 ) ) / 2 ) , ( 2nd𝑥 ) ⟩ ) ) )
3 ruc.4 𝐶 = ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ 𝐹 )
4 ruc.5 𝐺 = seq 0 ( 𝐷 , 𝐶 )
5 4 fveq1i ( 𝐺 ‘ 0 ) = ( seq 0 ( 𝐷 , 𝐶 ) ‘ 0 )
6 0z 0 ∈ ℤ
7 dfn2 ℕ = ( ℕ0 ∖ { 0 } )
8 7 reseq2i ( 𝐹 ↾ ℕ ) = ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) )
9 ffn ( 𝐹 : ℕ ⟶ ℝ → 𝐹 Fn ℕ )
10 fnresdm ( 𝐹 Fn ℕ → ( 𝐹 ↾ ℕ ) = 𝐹 )
11 1 9 10 3syl ( 𝜑 → ( 𝐹 ↾ ℕ ) = 𝐹 )
12 8 11 syl5reqr ( 𝜑𝐹 = ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) )
13 12 uneq2d ( 𝜑 → ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ 𝐹 ) = ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) )
14 3 13 syl5eq ( 𝜑𝐶 = ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) )
15 14 fveq1d ( 𝜑 → ( 𝐶 ‘ 0 ) = ( ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) ‘ 0 ) )
16 c0ex 0 ∈ V
17 16 a1i ( ⊤ → 0 ∈ V )
18 opex ⟨ 0 , 1 ⟩ ∈ V
19 18 a1i ( ⊤ → ⟨ 0 , 1 ⟩ ∈ V )
20 eqid ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) = ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) )
21 17 19 20 fvsnun1 ( ⊤ → ( ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) ‘ 0 ) = ⟨ 0 , 1 ⟩ )
22 21 mptru ( ( { ⟨ 0 , ⟨ 0 , 1 ⟩ ⟩ } ∪ ( 𝐹 ↾ ( ℕ0 ∖ { 0 } ) ) ) ‘ 0 ) = ⟨ 0 , 1 ⟩
23 15 22 eqtrdi ( 𝜑 → ( 𝐶 ‘ 0 ) = ⟨ 0 , 1 ⟩ )
24 6 23 seq1i ( 𝜑 → ( seq 0 ( 𝐷 , 𝐶 ) ‘ 0 ) = ⟨ 0 , 1 ⟩ )
25 5 24 syl5eq ( 𝜑 → ( 𝐺 ‘ 0 ) = ⟨ 0 , 1 ⟩ )