Metamath Proof Explorer


Theorem itcovalpc

Description: The value of the function that returns the n-th iterate of the "plus a constant" function with regard to composition. (Contributed by AV, 4-May-2024)

Ref Expression
Hypothesis itcovalpc.f ⊢ F = n ∈ ℕ 0 ⟼ n + C
Assertion itcovalpc ⊢ I ∈ ℕ 0 ∧ C ∈ ℕ 0 → IterComp ⁡ F ⁡ I = n ∈ ℕ 0 ⟼ n + C ⋅ I

Proof

Step Hyp Ref Expression
1 itcovalpc.f ⊢ F = n ∈ ℕ 0 ⟼ n + C
2 fveq2 ⊢ x = 0 → IterComp ⁡ F ⁡ x = IterComp ⁡ F ⁡ 0
3 oveq2 ⊢ x = 0 → C ⁢ x = C ⋅ 0
4 3 oveq2d ⊢ x = 0 → n + C ⁢ x = n + C ⋅ 0
5 4 mpteq2dv ⊢ x = 0 → n ∈ ℕ 0 ⟼ n + C ⁢ x = n ∈ ℕ 0 ⟼ n + C ⋅ 0
6 2 5 eqeq12d ⊢ x = 0 → IterComp ⁡ F ⁡ x = n ∈ ℕ 0 ⟼ n + C ⁢ x ↔ IterComp ⁡ F ⁡ 0 = n ∈ ℕ 0 ⟼ n + C ⋅ 0
7 fveq2 ⊢ x = y → IterComp ⁡ F ⁡ x = IterComp ⁡ F ⁡ y
8 oveq2 ⊢ x = y → C ⁢ x = C ⁢ y
9 8 oveq2d ⊢ x = y → n + C ⁢ x = n + C ⁢ y
10 9 mpteq2dv ⊢ x = y → n ∈ ℕ 0 ⟼ n + C ⁢ x = n ∈ ℕ 0 ⟼ n + C ⁢ y
11 7 10 eqeq12d ⊢ x = y → IterComp ⁡ F ⁡ x = n ∈ ℕ 0 ⟼ n + C ⁢ x ↔ IterComp ⁡ F ⁡ y = n ∈ ℕ 0 ⟼ n + C ⁢ y
12 fveq2 ⊢ x = y + 1 → IterComp ⁡ F ⁡ x = IterComp ⁡ F ⁡ y + 1
13 oveq2 ⊢ x = y + 1 → C ⁢ x = C ⁢ y + 1
14 13 oveq2d ⊢ x = y + 1 → n + C ⁢ x = n + C ⁢ y + 1
15 14 mpteq2dv ⊢ x = y + 1 → n ∈ ℕ 0 ⟼ n + C ⁢ x = n ∈ ℕ 0 ⟼ n + C ⁢ y + 1
16 12 15 eqeq12d ⊢ x = y + 1 → IterComp ⁡ F ⁡ x = n ∈ ℕ 0 ⟼ n + C ⁢ x ↔ IterComp ⁡ F ⁡ y + 1 = n ∈ ℕ 0 ⟼ n + C ⁢ y + 1
17 fveq2 ⊢ x = I → IterComp ⁡ F ⁡ x = IterComp ⁡ F ⁡ I
18 oveq2 ⊢ x = I → C ⁢ x = C ⋅ I
19 18 oveq2d ⊢ x = I → n + C ⁢ x = n + C ⋅ I
20 19 mpteq2dv ⊢ x = I → n ∈ ℕ 0 ⟼ n + C ⁢ x = n ∈ ℕ 0 ⟼ n + C ⋅ I
21 17 20 eqeq12d ⊢ x = I → IterComp ⁡ F ⁡ x = n ∈ ℕ 0 ⟼ n + C ⁢ x ↔ IterComp ⁡ F ⁡ I = n ∈ ℕ 0 ⟼ n + C ⋅ I
22 1 itcovalpclem1 ⊢ C ∈ ℕ 0 → IterComp ⁡ F ⁡ 0 = n ∈ ℕ 0 ⟼ n + C ⋅ 0
23 1 itcovalpclem2 ⊢ y ∈ ℕ 0 ∧ C ∈ ℕ 0 → IterComp ⁡ F ⁡ y = n ∈ ℕ 0 ⟼ n + C ⁢ y → IterComp ⁡ F ⁡ y + 1 = n ∈ ℕ 0 ⟼ n + C ⁢ y + 1
24 23 ancoms ⊢ C ∈ ℕ 0 ∧ y ∈ ℕ 0 → IterComp ⁡ F ⁡ y = n ∈ ℕ 0 ⟼ n + C ⁢ y → IterComp ⁡ F ⁡ y + 1 = n ∈ ℕ 0 ⟼ n + C ⁢ y + 1
25 24 imp ⊢ C ∈ ℕ 0 ∧ y ∈ ℕ 0 ∧ IterComp ⁡ F ⁡ y = n ∈ ℕ 0 ⟼ n + C ⁢ y → IterComp ⁡ F ⁡ y + 1 = n ∈ ℕ 0 ⟼ n + C ⁢ y + 1
26 6 11 16 21 22 25 nn0indd ⊢ C ∈ ℕ 0 ∧ I ∈ ℕ 0 → IterComp ⁡ F ⁡ I = n ∈ ℕ 0 ⟼ n + C ⋅ I
27 26 ancoms ⊢ I ∈ ℕ 0 ∧ C ∈ ℕ 0 → IterComp ⁡ F ⁡ I = n ∈ ℕ 0 ⟼ n + C ⋅ I