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Proof step 5 from proof square-nls (previous step)

sqrt(j) < i, i ^2 ≤ j, 0 ≤ j → sqrt(j) ^2 ≤ j ∧ j < (sqrt(j) + 1)^2 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 6:
0 ≤ j, sqrt(j) < i, i ^2 ≤ j, 0 ≤ j → sqrt(j) ^2 ≤ j ∧ j < (sqrt(j) + 1)^2 ⊦
Proof step 9:
sqrt(j) < i, i ^2 ≤ j, 0 ≤ j → sqrt(j) ^2 ≤ j ∧ j < (sqrt(j) + 1)^2 ⊦ 0 ≤ j

Rule argument

fma: 0 ≤ j

Additional information

Used formulas: 0 ≤ j