Metamath Proof Explorer


Theorem sqrtmsqd

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

Ref Expression
Hypotheses resqrcld.1 φA
resqrcld.2 φ0A
Assertion sqrtmsqd φAA=A

Proof

Step Hyp Ref Expression
1 resqrcld.1 φA
2 resqrcld.2 φ0A
3 sqrtmsq A0AAA=A
4 1 2 3 syl2anc φAA=A