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Proof step 14 from proof le-pos (previous step)

j0 = 0, 2 ≤ i, 1 < i, ~2 < i0, 0 ≤ i0, ~1 ≤ i0, j0 ≤ 0, j0 ≤ i0, ∀ 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 ≤ i ^ j0 ↔ i0 + 1 ≤ j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
2 ≤ i, 1 < i, ~2 < i0, ~1 ≤ i0, 0 ≤ i0, ∀ j1, i, j0. i→n(j1) < i→n(i0 +1) → 2 ≤ i ∧ 0 ≤ j1 ∧ 1 < i → (i ^ j1 ≤ i ^ j0 ↔ j1 ≤ j0) ⊦ i ^ i0 ≤ 0 ↔ i0 + 1 ≤ 0

Rule argument

Additional information

Used simplifier rules: Pot-zero: ⊦ j ^ 0 = 1;%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%le-one: ⊦ 1 < j → (i * j ≤ 1 ↔ i ≤ 0); from specification int-mult