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Proof step 2 from proof mod-nle (previous step)

i0 < i, 0 < j, 0 < i, 0 ≤ i0 ⊦ j * i + i0 ≤ j ↔ i = 1 ∧ i0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i0 < i, 0 < j, 0 < i, j * i + i0 ≤ j, 0 ≤ i0 ⊦ i = 1 ∧ i0 = 0

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 * i = i;%associativity: ⊦ i * j * j0 = (i * j) * j0;%commutativity: ⊦ i * j = j * i;%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs% ⊦ i ≤ i; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)