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Proof step 4 from proof mod2-2pot-02 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
i = i0 +1, 0 < i, j % (2 ^ i0 * 2) = 1, 0 < i, 0 ≤ j ⊦ j % 2 = 1

Rule argument

spec: int-basic1 inst: name: elim-pred seq: ⊦ i = j -1 ↔ j = i +1 subst: [j, i] → [i, i0] flags:

Additional information

Used simplifier rules: ⊦ i = j -1 ↔ j = i +1; (elim) Used terms: i, i0