Metamath Proof Explorer


Theorem expcllem

Description: Lemma for proving nonnegative integer exponentiation closure laws. (Contributed by NM, 14-Dec-2005)

Ref Expression
Hypotheses expcllem.1 ⊢ F ⊆ ℂ
expcllem.2 ⊢ x ∈ F ∧ y ∈ F → x ⁢ y ∈ F
expcllem.3 ⊢ 1 ∈ F
Assertion expcllem ⊢ A ∈ F ∧ B ∈ ℕ 0 → A B ∈ F

Proof

Step Hyp Ref Expression
1 expcllem.1 ⊢ F ⊆ ℂ
2 expcllem.2 ⊢ x ∈ F ∧ y ∈ F → x ⁢ y ∈ F
3 expcllem.3 ⊢ 1 ∈ F
4 elnn0 ⊢ B ∈ ℕ 0 ↔ B ∈ ℕ ∨ B = 0
5 oveq2 ⊢ z = 1 → A z = A 1
6 5 eleq1d ⊢ z = 1 → A z ∈ F ↔ A 1 ∈ F
7 6 imbi2d ⊢ z = 1 → A ∈ F → A z ∈ F ↔ A ∈ F → A 1 ∈ F
8 oveq2 ⊢ z = w → A z = A w
9 8 eleq1d ⊢ z = w → A z ∈ F ↔ A w ∈ F
10 9 imbi2d ⊢ z = w → A ∈ F → A z ∈ F ↔ A ∈ F → A w ∈ F
11 oveq2 ⊢ z = w + 1 → A z = A w + 1
12 11 eleq1d ⊢ z = w + 1 → A z ∈ F ↔ A w + 1 ∈ F
13 12 imbi2d ⊢ z = w + 1 → A ∈ F → A z ∈ F ↔ A ∈ F → A w + 1 ∈ F
14 oveq2 ⊢ z = B → A z = A B
15 14 eleq1d ⊢ z = B → A z ∈ F ↔ A B ∈ F
16 15 imbi2d ⊢ z = B → A ∈ F → A z ∈ F ↔ A ∈ F → A B ∈ F
17 1 sseli ⊢ A ∈ F → A ∈ ℂ
18 exp1 ⊢ A ∈ ℂ → A 1 = A
19 17 18 syl ⊢ A ∈ F → A 1 = A
20 19 eleq1d ⊢ A ∈ F → A 1 ∈ F ↔ A ∈ F
21 20 ibir ⊢ A ∈ F → A 1 ∈ F
22 2 caovcl ⊢ A w ∈ F ∧ A ∈ F → A w ⁢ A ∈ F
23 22 ancoms ⊢ A ∈ F ∧ A w ∈ F → A w ⁢ A ∈ F
24 23 adantlr ⊢ A ∈ F ∧ w ∈ ℕ ∧ A w ∈ F → A w ⁢ A ∈ F
25 nnnn0 ⊢ w ∈ ℕ → w ∈ ℕ 0
26 expp1 ⊢ A ∈ ℂ ∧ w ∈ ℕ 0 → A w + 1 = A w ⁢ A
27 17 25 26 syl2an ⊢ A ∈ F ∧ w ∈ ℕ → A w + 1 = A w ⁢ A
28 27 eleq1d ⊢ A ∈ F ∧ w ∈ ℕ → A w + 1 ∈ F ↔ A w ⁢ A ∈ F
29 28 adantr ⊢ A ∈ F ∧ w ∈ ℕ ∧ A w ∈ F → A w + 1 ∈ F ↔ A w ⁢ A ∈ F
30 24 29 mpbird ⊢ A ∈ F ∧ w ∈ ℕ ∧ A w ∈ F → A w + 1 ∈ F
31 30 exp31 ⊢ A ∈ F → w ∈ ℕ → A w ∈ F → A w + 1 ∈ F
32 31 com12 ⊢ w ∈ ℕ → A ∈ F → A w ∈ F → A w + 1 ∈ F
33 32 a2d ⊢ w ∈ ℕ → A ∈ F → A w ∈ F → A ∈ F → A w + 1 ∈ F
34 7 10 13 16 21 33 nnind ⊢ B ∈ ℕ → A ∈ F → A B ∈ F
35 34 impcom ⊢ A ∈ F ∧ B ∈ ℕ → A B ∈ F
36 oveq2 ⊢ B = 0 → A B = A 0
37 exp0 ⊢ A ∈ ℂ → A 0 = 1
38 17 37 syl ⊢ A ∈ F → A 0 = 1
39 36 38 sylan9eqr ⊢ A ∈ F ∧ B = 0 → A B = 1
40 39 3 eqeltrdi ⊢ A ∈ F ∧ B = 0 → A B ∈ F
41 35 40 jaodan ⊢ A ∈ F ∧ B ∈ ℕ ∨ B = 0 → A B ∈ F
42 4 41 sylan2b ⊢ A ∈ F ∧ B ∈ ℕ 0 → A B ∈ F