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Proof step 18 from proof fac-le (previous step)

(i1 + 1) * fac(i1) ≤ (i2 + 1) * fac(i2), fac(i1) ≤ fac(i2) ↔ i1 ≤ i2, 0 ≤ i1, 0 ≤ i2, ∀ i0, j. i→n(i0) < i→n(i1 + 1) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ i1 ≤ i2

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
(i1 + 1) * fac(i1) ≤ (i2 + 1) * fac(i2), 0 ≤ i1, 0 ≤ i2, i2 < i1, fac(i2) < fac(i1), 0 < i1, ∀ i0, j. i→n(i0) < i→n(i1 + 1) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs