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Proof step 19 from proof le-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: le-def seq: ⊦ i ≤ j ↔ ¬ j < i subst: [j, i] → [i0, i0 + 1] flags: <all paths>

Additional information

Used simplifier rules: i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j; (s)% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; (s)% ⊦ i ≤ ~1 ↔ i < 0; (s)% ⊦ i ≤ j → ¬ j < i; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: i0, i0 + 1