Metamath Proof Explorer


Theorem sqrtmsqd

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

Ref Expression
Hypotheses resqrcld.1 φ A
resqrcld.2 φ 0 A
Assertion sqrtmsqd φ A A = A

Proof

Step Hyp Ref Expression
1 resqrcld.1 φ A
2 resqrcld.2 φ 0 A
3 sqrtmsq A 0 A A A = A
4 1 2 3 syl2anc φ A A = A