Metamath Proof Explorer


Theorem cnsrexpcl

Description: Exponentiation is closed in number rings. (Contributed by Stefan O'Rear, 30-Nov-2014)

Ref Expression
Hypotheses cnsrexpcl.s ⊢ φ → S ∈ SubRing ⁡ ℂ fld
cnsrexpcl.x ⊢ φ → X ∈ S
cnsrexpcl.y ⊢ φ → Y ∈ ℕ 0
Assertion cnsrexpcl ⊢ φ → X Y ∈ S

Proof

Step Hyp Ref Expression
1 cnsrexpcl.s ⊢ φ → S ∈ SubRing ⁡ ℂ fld
2 cnsrexpcl.x ⊢ φ → X ∈ S
3 cnsrexpcl.y ⊢ φ → Y ∈ ℕ 0
4 oveq2 ⊢ a = 0 → X a = X 0
5 4 eleq1d ⊢ a = 0 → X a ∈ S ↔ X 0 ∈ S
6 5 imbi2d ⊢ a = 0 → φ → X a ∈ S ↔ φ → X 0 ∈ S
7 oveq2 ⊢ a = b → X a = X b
8 7 eleq1d ⊢ a = b → X a ∈ S ↔ X b ∈ S
9 8 imbi2d ⊢ a = b → φ → X a ∈ S ↔ φ → X b ∈ S
10 oveq2 ⊢ a = b + 1 → X a = X b + 1
11 10 eleq1d ⊢ a = b + 1 → X a ∈ S ↔ X b + 1 ∈ S
12 11 imbi2d ⊢ a = b + 1 → φ → X a ∈ S ↔ φ → X b + 1 ∈ S
13 oveq2 ⊢ a = Y → X a = X Y
14 13 eleq1d ⊢ a = Y → X a ∈ S ↔ X Y ∈ S
15 14 imbi2d ⊢ a = Y → φ → X a ∈ S ↔ φ → X Y ∈ S
16 cnfldbas ⊢ ℂ = Base ℂ fld
17 16 subrgss ⊢ S ∈ SubRing ⁡ ℂ fld → S ⊆ ℂ
18 1 17 syl ⊢ φ → S ⊆ ℂ
19 18 2 sseldd ⊢ φ → X ∈ ℂ
20 19 exp0d ⊢ φ → X 0 = 1
21 cnfld1 ⊢ 1 = 1 ℂ fld
22 21 subrg1cl ⊢ S ∈ SubRing ⁡ ℂ fld → 1 ∈ S
23 1 22 syl ⊢ φ → 1 ∈ S
24 20 23 eqeltrd ⊢ φ → X 0 ∈ S
25 19 3ad2ant2 ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X ∈ ℂ
26 simp1 ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → b ∈ ℕ 0
27 25 26 expp1d ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X b + 1 = X b ⁢ X
28 1 3ad2ant2 ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → S ∈ SubRing ⁡ ℂ fld
29 simp3 ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X b ∈ S
30 2 3ad2ant2 ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X ∈ S
31 cnfldmul ⊢ × = ⋅ ℂ fld
32 31 subrgmcl ⊢ S ∈ SubRing ⁡ ℂ fld ∧ X b ∈ S ∧ X ∈ S → X b ⁢ X ∈ S
33 28 29 30 32 syl3anc ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X b ⁢ X ∈ S
34 27 33 eqeltrd ⊢ b ∈ ℕ 0 ∧ φ ∧ X b ∈ S → X b + 1 ∈ S
35 34 3exp ⊢ b ∈ ℕ 0 → φ → X b ∈ S → X b + 1 ∈ S
36 35 a2d ⊢ b ∈ ℕ 0 → φ → X b ∈ S → φ → X b + 1 ∈ S
37 6 9 12 15 24 36 nn0ind ⊢ Y ∈ ℕ 0 → φ → X Y ∈ S
38 3 37 mpcom ⊢ φ → X Y ∈ S