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Proof step 6 from proof fac-le (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
i = i1 + 1, 0 < i, 0 < j, 0 < i, ∀ i0, j. i→n(i0) < i→n(i) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ i * fac(i1) ≤ fac(j) ↔ i ≤ j

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i, 1, i1] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i, 1, i1