Metamath Proof Explorer


Theorem fltdiv

Description: A counterexample to FLT stays valid when scaled. The hypotheses are more general than they need to be for convenience. (Contributed by SN, 20-Aug-2024)

Ref Expression
Hypotheses fltdiv.s ⊢ φ → S ∈ ℂ
fltdiv.0 ⊢ φ → S ≠ 0
fltdiv.a ⊢ φ → A ∈ ℂ
fltdiv.b ⊢ φ → B ∈ ℂ
fltdiv.c ⊢ φ → C ∈ ℂ
fltdiv.n ⊢ φ → N ∈ ℕ 0
fltdiv.1 ⊢ φ → A N + B N = C N
Assertion fltdiv ⊢ φ → A S N + B S N = C S N

Proof

Step Hyp Ref Expression
1 fltdiv.s ⊢ φ → S ∈ ℂ
2 fltdiv.0 ⊢ φ → S ≠ 0
3 fltdiv.a ⊢ φ → A ∈ ℂ
4 fltdiv.b ⊢ φ → B ∈ ℂ
5 fltdiv.c ⊢ φ → C ∈ ℂ
6 fltdiv.n ⊢ φ → N ∈ ℕ 0
7 fltdiv.1 ⊢ φ → A N + B N = C N
8 3 6 expcld ⊢ φ → A N ∈ ℂ
9 4 6 expcld ⊢ φ → B N ∈ ℂ
10 1 6 expcld ⊢ φ → S N ∈ ℂ
11 6 nn0zd ⊢ φ → N ∈ ℤ
12 1 2 11 expne0d ⊢ φ → S N ≠ 0
13 8 9 10 12 divdird ⊢ φ → A N + B N S N = A N S N + B N S N
14 7 oveq1d ⊢ φ → A N + B N S N = C N S N
15 13 14 eqtr3d ⊢ φ → A N S N + B N S N = C N S N
16 3 1 2 6 expdivd ⊢ φ → A S N = A N S N
17 4 1 2 6 expdivd ⊢ φ → B S N = B N S N
18 16 17 oveq12d ⊢ φ → A S N + B S N = A N S N + B N S N
19 5 1 2 6 expdivd ⊢ φ → C S N = C N S N
20 15 18 19 3eqtr4d ⊢ φ → A S N + B S N = C S N