Metamath Proof Explorer


Theorem sqrtsqi

Description: Square root of square. (Contributed by NM, 11-Aug-1999)

Ref Expression
Hypothesis sqrtthi.1
|- A e. RR
Assertion sqrtsqi
|- ( 0 <_ A -> ( sqrt ` ( A ^ 2 ) ) = A )

Proof

Step Hyp Ref Expression
1 sqrtthi.1
 |-  A e. RR
2 sqrtsq
 |-  ( ( A e. RR /\ 0 <_ A ) -> ( sqrt ` ( A ^ 2 ) ) = A )
3 1 2 mpan
 |-  ( 0 <_ A -> ( sqrt ` ( A ^ 2 ) ) = A )