Metamath Proof Explorer


Theorem vdwapval

Description: Value of the arithmetic progression function. (Contributed by Mario Carneiro, 18-Aug-2014)

Ref Expression
Assertion vdwapval ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → X ∈ A AP ⁡ K D ↔ ∃ m ∈ 0 … K − 1 X = A + m ⁢ D

Proof

Step Hyp Ref Expression
1 vdwapfval ⊢ K ∈ ℕ 0 → AP ⁡ K = a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d
2 1 3ad2ant1 ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → AP ⁡ K = a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d
3 2 oveqd ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → A AP ⁡ K D = A a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d D
4 oveq2 ⊢ d = D → m ⁢ d = m ⁢ D
5 oveq12 ⊢ a = A ∧ m ⁢ d = m ⁢ D → a + m ⁢ d = A + m ⁢ D
6 4 5 sylan2 ⊢ a = A ∧ d = D → a + m ⁢ d = A + m ⁢ D
7 6 mpteq2dv ⊢ a = A ∧ d = D → m ∈ 0 … K − 1 ⟼ a + m ⁢ d = m ∈ 0 … K − 1 ⟼ A + m ⁢ D
8 7 rneqd ⊢ a = A ∧ d = D → ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d = ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D
9 eqid ⊢ a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d = a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d
10 ovex ⊢ 0 … K − 1 ∈ V
11 10 mptex ⊢ m ∈ 0 … K − 1 ⟼ A + m ⁢ D ∈ V
12 11 rnex ⊢ ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D ∈ V
13 8 9 12 ovmpoa ⊢ A ∈ ℕ ∧ D ∈ ℕ → A a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d D = ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D
14 13 3adant1 ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → A a ∈ ℕ , d ∈ ℕ ⟼ ran ⁡ m ∈ 0 … K − 1 ⟼ a + m ⁢ d D = ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D
15 3 14 eqtrd ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → A AP ⁡ K D = ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D
16 eqid ⊢ m ∈ 0 … K − 1 ⟼ A + m ⁢ D = m ∈ 0 … K − 1 ⟼ A + m ⁢ D
17 16 rnmpt ⊢ ran ⁡ m ∈ 0 … K − 1 ⟼ A + m ⁢ D = x | ∃ m ∈ 0 … K − 1 x = A + m ⁢ D
18 15 17 eqtrdi ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → A AP ⁡ K D = x | ∃ m ∈ 0 … K − 1 x = A + m ⁢ D
19 18 eleq2d ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → X ∈ A AP ⁡ K D ↔ X ∈ x | ∃ m ∈ 0 … K − 1 x = A + m ⁢ D
20 id ⊢ X = A + m ⁢ D → X = A + m ⁢ D
21 ovex ⊢ A + m ⁢ D ∈ V
22 20 21 eqeltrdi ⊢ X = A + m ⁢ D → X ∈ V
23 22 rexlimivw ⊢ ∃ m ∈ 0 … K − 1 X = A + m ⁢ D → X ∈ V
24 eqeq1 ⊢ x = X → x = A + m ⁢ D ↔ X = A + m ⁢ D
25 24 rexbidv ⊢ x = X → ∃ m ∈ 0 … K − 1 x = A + m ⁢ D ↔ ∃ m ∈ 0 … K − 1 X = A + m ⁢ D
26 23 25 elab3 ⊢ X ∈ x | ∃ m ∈ 0 … K − 1 x = A + m ⁢ D ↔ ∃ m ∈ 0 … K − 1 X = A + m ⁢ D
27 19 26 bitrdi ⊢ K ∈ ℕ 0 ∧ A ∈ ℕ ∧ D ∈ ℕ → X ∈ A AP ⁡ K D ↔ ∃ m ∈ 0 … K − 1 X = A + m ⁢ D