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Proof step 35 from proof le-pos (previous step)

0 < i, ~1 ≤ i0, 0 ≤ i0, ~2 < i0, 1 < i, 2 ≤ i, 0 ≤ i1, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 ≤ i ^ i1 ↔ i0 < i1 + 1

User applied rule apply induction

And obtained the following new goals

Rule argument

subst: [j1, i, j0] → [i0, i, i1]

Additional information

Used simplifier rules: succ-01: ⊦ i +1 = i + 1; from specification int-basic2%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%irref: ⊦ ¬ i < i; from specification int-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: i0, i, i1