Metamath Proof Explorer


Theorem fltabcoprmex

Description: A counterexample to FLT implies a counterexample to FLT with A , B (assigned to A / ( A gcd B ) and B / ( A gcd B ) ) coprime (by divgcdcoprm0 ). (Contributed by SN, 20-Aug-2024)

Ref Expression
Hypotheses fltabcoprmex.a ⊢ φ → A ∈ ℕ
fltabcoprmex.b ⊢ φ → B ∈ ℕ
fltabcoprmex.c ⊢ φ → C ∈ ℕ
fltabcoprmex.n ⊢ φ → N ∈ ℕ 0
fltabcoprmex.1 ⊢ φ → A N + B N = C N
Assertion fltabcoprmex ⊢ φ → A A gcd B N + B A gcd B N = C A gcd B N

Proof

Step Hyp Ref Expression
1 fltabcoprmex.a ⊢ φ → A ∈ ℕ
2 fltabcoprmex.b ⊢ φ → B ∈ ℕ
3 fltabcoprmex.c ⊢ φ → C ∈ ℕ
4 fltabcoprmex.n ⊢ φ → N ∈ ℕ 0
5 fltabcoprmex.1 ⊢ φ → A N + B N = C N
6 gcdnncl ⊢ A ∈ ℕ ∧ B ∈ ℕ → A gcd B ∈ ℕ
7 1 2 6 syl2anc ⊢ φ → A gcd B ∈ ℕ
8 7 nncnd ⊢ φ → A gcd B ∈ ℂ
9 7 nnne0d ⊢ φ → A gcd B ≠ 0
10 1 nncnd ⊢ φ → A ∈ ℂ
11 2 nncnd ⊢ φ → B ∈ ℂ
12 3 nncnd ⊢ φ → C ∈ ℂ
13 8 9 10 11 12 4 5 fltdiv ⊢ φ → A A gcd B N + B A gcd B N = C A gcd B N