Metamath Proof Explorer


Theorem itcovalt2lem1

Description: Lemma 1 for itcovalt2 : induction basis. (Contributed by AV, 5-May-2024)

Ref Expression
Hypothesis itcovalt2.f 𝐹 = ( 𝑛 ∈ ℕ0 ↦ ( ( 2 · 𝑛 ) + 𝐶 ) )
Assertion itcovalt2lem1 ( 𝐶 ∈ ℕ0 → ( ( IterComp ‘ 𝐹 ) ‘ 0 ) = ( 𝑛 ∈ ℕ0 ↦ ( ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) − 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 itcovalt2.f 𝐹 = ( 𝑛 ∈ ℕ0 ↦ ( ( 2 · 𝑛 ) + 𝐶 ) )
2 nn0ex 0 ∈ V
3 ovexd ( 𝑛 ∈ ℕ0 → ( ( 2 · 𝑛 ) + 𝐶 ) ∈ V )
4 3 rgen 𝑛 ∈ ℕ0 ( ( 2 · 𝑛 ) + 𝐶 ) ∈ V
5 2 4 pm3.2i ( ℕ0 ∈ V ∧ ∀ 𝑛 ∈ ℕ0 ( ( 2 · 𝑛 ) + 𝐶 ) ∈ V )
6 1 itcoval0mpt ( ( ℕ0 ∈ V ∧ ∀ 𝑛 ∈ ℕ0 ( ( 2 · 𝑛 ) + 𝐶 ) ∈ V ) → ( ( IterComp ‘ 𝐹 ) ‘ 0 ) = ( 𝑛 ∈ ℕ0𝑛 ) )
7 5 6 mp1i ( 𝐶 ∈ ℕ0 → ( ( IterComp ‘ 𝐹 ) ‘ 0 ) = ( 𝑛 ∈ ℕ0𝑛 ) )
8 simpr ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → 𝑛 ∈ ℕ0 )
9 8 nn0cnd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → 𝑛 ∈ ℂ )
10 simpl ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → 𝐶 ∈ ℕ0 )
11 10 nn0cnd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → 𝐶 ∈ ℂ )
12 2nn0 2 ∈ ℕ0
13 12 numexp0 ( 2 ↑ 0 ) = 1
14 13 a1i ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( 2 ↑ 0 ) = 1 )
15 14 oveq2d ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) = ( ( 𝑛 + 𝐶 ) · 1 ) )
16 8 10 nn0addcld ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( 𝑛 + 𝐶 ) ∈ ℕ0 )
17 16 nn0cnd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( 𝑛 + 𝐶 ) ∈ ℂ )
18 17 mulid1d ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( ( 𝑛 + 𝐶 ) · 1 ) = ( 𝑛 + 𝐶 ) )
19 15 18 eqtrd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) = ( 𝑛 + 𝐶 ) )
20 9 11 19 mvrraddd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → ( ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) − 𝐶 ) = 𝑛 )
21 20 eqcomd ( ( 𝐶 ∈ ℕ0𝑛 ∈ ℕ0 ) → 𝑛 = ( ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) − 𝐶 ) )
22 21 mpteq2dva ( 𝐶 ∈ ℕ0 → ( 𝑛 ∈ ℕ0𝑛 ) = ( 𝑛 ∈ ℕ0 ↦ ( ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) − 𝐶 ) ) )
23 7 22 eqtrd ( 𝐶 ∈ ℕ0 → ( ( IterComp ‘ 𝐹 ) ‘ 0 ) = ( 𝑛 ∈ ℕ0 ↦ ( ( ( 𝑛 + 𝐶 ) · ( 2 ↑ 0 ) ) − 𝐶 ) ) )