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Proof step 12 from proof fac-le (previous step)

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) ↔ i1 ≤ i2

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [i0, j] → [i1, i2]

Additional information

Used simplifier rules: lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%rax-04: ⊦ 0 ≤ i → i→n(i + 1) = i→n(i) + 1; from specification intnat%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%labs: ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j); from specification intnat%absle: ⊦ 0 ≤ i → abs(i) = i; from specification int-abs%irreflexivity: ⊦ ¬ i < i; from specification oelem[intlista]%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i1, i2