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Proof step 21 from proof fac-le (previous step)

i1 ≤ 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 + 1) * fac(i1) ≤ (i2 + 1) * fac(i2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
i1 ≤ i2, 0 ≤ i1, 0 ≤ i2, fac(i1) ≤ fac(i2), (i2 + 1) * fac(i2) < (i1 + 1) * fac(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: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs