Metamath Proof Explorer


Theorem sqrtsqd

Description: Square root of square. (Contributed by Mario Carneiro, 29-May-2016)

Ref Expression
Hypotheses resqrcld.1
|- ( ph -> A e. RR )
resqrcld.2
|- ( ph -> 0 <_ A )
Assertion sqrtsqd
|- ( ph -> ( sqrt ` ( A ^ 2 ) ) = A )

Proof

Step Hyp Ref Expression
1 resqrcld.1
 |-  ( ph -> A e. RR )
2 resqrcld.2
 |-  ( ph -> 0 <_ A )
3 sqrtsq
 |-  ( ( A e. RR /\ 0 <_ A ) -> ( sqrt ` ( A ^ 2 ) ) = A )
4 1 2 3 syl2anc
 |-  ( ph -> ( sqrt ` ( A ^ 2 ) ) = A )