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Proof step 7 from proof one-01 (previous step)

1 < (sqrt(1) + 1) * (sqrt(1) + 1), sqrt(1) * sqrt(1) ≤ 1, 0 < (sqrt(1) + 1) * (sqrt(1) + 1), sqrt(1) < abs(sqrt(1) + 1), sqrt(1) = 0 ∨ sqrt(1) = ~1 ⊦ sqrt(1) = 1

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 5

Additional information

Used simplifier rules: lf-03: ⊦ i < j ∧ i = i0 → i0 < j; (forward) from specification int-basic1%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlsf-01: ⊦ (* i) < j ∧ 0 < i → 0 ≤ j ∧ 0 < j; (forward) from specification int-abs%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs