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Proof step 7 from proof fac-le (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
0 < j, 0 ≤ i1, ∀ i0, j. i→n(i0) < i→n(i1 + 1) → 0 < i0 ∧ 0 < j → (fac(i0) ≤ fac(j) ↔ i0 ≤ j) ⊦ (i1 + 1) * fac(i1) ≤ fac(j) ↔ i1 < j

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%succ-01: ⊦ i + 1 ≤ j ↔ i < j; from specification int-abs