Metamath Proof Explorer


Theorem flt0

Description: A counterexample for FLT does not exist for N = 0 . (Contributed by SN, 20-Aug-2024)

Ref Expression
Hypotheses flt0.a ( 𝜑𝐴 ∈ ℂ )
flt0.b ( 𝜑𝐵 ∈ ℂ )
flt0.c ( 𝜑𝐶 ∈ ℂ )
flt0.n ( 𝜑𝑁 ∈ ℕ0 )
flt0.1 ( 𝜑 → ( ( 𝐴𝑁 ) + ( 𝐵𝑁 ) ) = ( 𝐶𝑁 ) )
Assertion flt0 ( 𝜑𝑁 ∈ ℕ )

Proof

Step Hyp Ref Expression
1 flt0.a ( 𝜑𝐴 ∈ ℂ )
2 flt0.b ( 𝜑𝐵 ∈ ℂ )
3 flt0.c ( 𝜑𝐶 ∈ ℂ )
4 flt0.n ( 𝜑𝑁 ∈ ℕ0 )
5 flt0.1 ( 𝜑 → ( ( 𝐴𝑁 ) + ( 𝐵𝑁 ) ) = ( 𝐶𝑁 ) )
6 1p1e2 ( 1 + 1 ) = 2
7 sn-1ne2 1 ≠ 2
8 7 necomi 2 ≠ 1
9 6 8 eqnetri ( 1 + 1 ) ≠ 1
10 9 a1i ( 𝜑 → ( 1 + 1 ) ≠ 1 )
11 1 exp0d ( 𝜑 → ( 𝐴 ↑ 0 ) = 1 )
12 2 exp0d ( 𝜑 → ( 𝐵 ↑ 0 ) = 1 )
13 11 12 oveq12d ( 𝜑 → ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) = ( 1 + 1 ) )
14 3 exp0d ( 𝜑 → ( 𝐶 ↑ 0 ) = 1 )
15 10 13 14 3netr4d ( 𝜑 → ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) ≠ ( 𝐶 ↑ 0 ) )
16 oveq2 ( 𝑁 = 0 → ( 𝐴𝑁 ) = ( 𝐴 ↑ 0 ) )
17 oveq2 ( 𝑁 = 0 → ( 𝐵𝑁 ) = ( 𝐵 ↑ 0 ) )
18 16 17 oveq12d ( 𝑁 = 0 → ( ( 𝐴𝑁 ) + ( 𝐵𝑁 ) ) = ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) )
19 oveq2 ( 𝑁 = 0 → ( 𝐶𝑁 ) = ( 𝐶 ↑ 0 ) )
20 18 19 eqeq12d ( 𝑁 = 0 → ( ( ( 𝐴𝑁 ) + ( 𝐵𝑁 ) ) = ( 𝐶𝑁 ) ↔ ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) = ( 𝐶 ↑ 0 ) ) )
21 5 20 syl5ibcom ( 𝜑 → ( 𝑁 = 0 → ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) = ( 𝐶 ↑ 0 ) ) )
22 21 imp ( ( 𝜑𝑁 = 0 ) → ( ( 𝐴 ↑ 0 ) + ( 𝐵 ↑ 0 ) ) = ( 𝐶 ↑ 0 ) )
23 15 22 mteqand ( 𝜑𝑁 ≠ 0 )
24 elnnne0 ( 𝑁 ∈ ℕ ↔ ( 𝑁 ∈ ℕ0𝑁 ≠ 0 ) )
25 4 23 24 sylanbrc ( 𝜑𝑁 ∈ ℕ )