Metamath Proof Explorer


Theorem sqrtsq

Description: Square root of square. (Contributed by NM, 14-Jan-2006) (Revised by Mario Carneiro, 29-May-2016)

Ref Expression
Assertion sqrtsq
|- ( ( A e. RR /\ 0 <_ A ) -> ( sqrt ` ( A ^ 2 ) ) = A )

Proof

Step Hyp Ref Expression
1 eqid
 |-  ( A ^ 2 ) = ( A ^ 2 )
2 resqcl
 |-  ( A e. RR -> ( A ^ 2 ) e. RR )
3 sqge0
 |-  ( A e. RR -> 0 <_ ( A ^ 2 ) )
4 2 3 jca
 |-  ( A e. RR -> ( ( A ^ 2 ) e. RR /\ 0 <_ ( A ^ 2 ) ) )
5 4 adantr
 |-  ( ( A e. RR /\ 0 <_ A ) -> ( ( A ^ 2 ) e. RR /\ 0 <_ ( A ^ 2 ) ) )
6 sqrtsq2
 |-  ( ( ( ( A ^ 2 ) e. RR /\ 0 <_ ( A ^ 2 ) ) /\ ( A e. RR /\ 0 <_ A ) ) -> ( ( sqrt ` ( A ^ 2 ) ) = A <-> ( A ^ 2 ) = ( A ^ 2 ) ) )
7 5 6 mpancom
 |-  ( ( A e. RR /\ 0 <_ A ) -> ( ( sqrt ` ( A ^ 2 ) ) = A <-> ( A ^ 2 ) = ( A ^ 2 ) ) )
8 1 7 mpbiri
 |-  ( ( A e. RR /\ 0 <_ A ) -> ( sqrt ` ( A ^ 2 ) ) = A )