Metamath Proof Explorer


Theorem resqrtthlem

Description: Lemma for resqrtth . (Contributed by Mario Carneiro, 9-Jul-2013)

Ref Expression
Assertion resqrtthlem A 0 A A 2 = A 0 A i A +

Proof

Step Hyp Ref Expression
1 recn A A
2 sqrtval A A = ι x | x 2 = A 0 x i x +
3 2 eqcomd A ι x | x 2 = A 0 x i x + = A
4 1 3 syl A ι x | x 2 = A 0 x i x + = A
5 4 adantr A 0 A ι x | x 2 = A 0 x i x + = A
6 resqrtcl A 0 A A
7 6 recnd A 0 A A
8 resqreu A 0 A ∃! x x 2 = A 0 x i x +
9 oveq1 x = A x 2 = A 2
10 9 eqeq1d x = A x 2 = A A 2 = A
11 fveq2 x = A x = A
12 11 breq2d x = A 0 x 0 A
13 oveq2 x = A i x = i A
14 neleq1 i x = i A i x + i A +
15 13 14 syl x = A i x + i A +
16 10 12 15 3anbi123d x = A x 2 = A 0 x i x + A 2 = A 0 A i A +
17 16 riota2 A ∃! x x 2 = A 0 x i x + A 2 = A 0 A i A + ι x | x 2 = A 0 x i x + = A
18 7 8 17 syl2anc A 0 A A 2 = A 0 A i A + ι x | x 2 = A 0 x i x + = A
19 5 18 mpbird A 0 A A 2 = A 0 A i A +