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Proof step 10 from proof le-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
~2 < i0, 1 < i, 2 ≤ i, 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) ⊦ i ^ i0 * i ≤ i ^ j0 ↔ i0 + 1 ≤ j0

Rule argument

Additional information

Used simplifier rules: numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ (* j) ≤ i0 + (* j0) ↔ j - j0 ≤ i0; (s)% ⊦ i +1 = i + 1; (s)