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 φ A
flt0.b φ B
flt0.c φ C
flt0.n φ N 0
flt0.1 φ A N + B N = C N
Assertion flt0 φ N

Proof

Step Hyp Ref Expression
1 flt0.a φ A
2 flt0.b φ B
3 flt0.c φ C
4 flt0.n φ N 0
5 flt0.1 φ A N + B N = C N
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 A N = A 0
17 oveq2 N = 0 B N = B 0
18 16 17 oveq12d N = 0 A N + B N = A 0 + B 0
19 oveq2 N = 0 C N = C 0
20 18 19 eqeq12d N = 0 A N + B N = C N A 0 + B 0 = C 0
21 5 20 syl5ibcom φ N = 0 A 0 + B 0 = C 0
22 21 imp φ N = 0 A 0 + B 0 = C 0
23 15 22 mteqand φ N 0
24 elnnne0 N N 0 N 0
25 4 23 24 sylanbrc φ N