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Proof step 6 from proof mod2-2pot-02 (previous step)

j % (2 ^ i0 * 2) = 1, i = i0 +1, 0 < i, 0 ≤ j ⊦ j % 2 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
j % (2 ^ i0 * 2) = 1, 0 ≤ j, 0 ≤ i0 ⊦ j % 2 = 1

Rule argument

Additional information

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