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Proof step 7 from proof fac-not-neg (previous step)

0 < i, fac(i - 1) < 0, fac(i - 1) < i, ∀ i0. i→n(i0) < i→n(i) → fac(i0) < 0 → false ⊦

User applied rule apply induction

And closed the goal!

Rule argument

subst: [i0] → [i - 1]

Additional information

Used simplifier rules: abs-pred: ⊦ i→n(i - 1) < i→n(i) ↔ 0 < i; from specification intnat%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: i - 1