Metamath Proof Explorer


Theorem jm2.21

Description: Lemma for jm2.20nn . Express X and Y values as a binomial. (Contributed by Stefan O'Rear, 26-Sep-2014)

Ref Expression
Assertion jm2.21 ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N ⋅ J + A 2 − 1 ⁢ A Y rm N ⋅ J = A X rm N + A 2 − 1 ⁢ A Y rm N J

Proof

Step Hyp Ref Expression
1 rmbaserp ⊢ A ∈ ℤ ≥ 2 → A + A 2 − 1 ∈ ℝ +
2 1 rpcnne0d ⊢ A ∈ ℤ ≥ 2 → A + A 2 − 1 ∈ ℂ ∧ A + A 2 − 1 ≠ 0
3 expmulz ⊢ A + A 2 − 1 ∈ ℂ ∧ A + A 2 − 1 ≠ 0 ∧ N ∈ ℤ ∧ J ∈ ℤ → A + A 2 − 1 N ⋅ J = A + A 2 − 1 N J
4 2 3 sylan ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A + A 2 − 1 N ⋅ J = A + A 2 − 1 N J
5 zmulcl ⊢ N ∈ ℤ ∧ J ∈ ℤ → N ⋅ J ∈ ℤ
6 rmxyval ⊢ A ∈ ℤ ≥ 2 ∧ N ⋅ J ∈ ℤ → A X rm N ⋅ J + A 2 − 1 ⁢ A Y rm N ⋅ J = A + A 2 − 1 N ⋅ J
7 5 6 sylan2 ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N ⋅ J + A 2 − 1 ⁢ A Y rm N ⋅ J = A + A 2 − 1 N ⋅ J
8 rmxyval ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N = A + A 2 − 1 N
9 8 adantrr ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N = A + A 2 − 1 N
10 9 oveq1d ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N + A 2 − 1 ⁢ A Y rm N J = A + A 2 − 1 N J
11 4 7 10 3eqtr4d ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N ⋅ J + A 2 − 1 ⁢ A Y rm N ⋅ J = A X rm N + A 2 − 1 ⁢ A Y rm N J
12 11 3impb ⊢ A ∈ ℤ ≥ 2 ∧ N ∈ ℤ ∧ J ∈ ℤ → A X rm N ⋅ J + A 2 − 1 ⁢ A Y rm N ⋅ J = A X rm N + A 2 − 1 ⁢ A Y rm N J