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Proof step 5 from proof div2 (previous step)

⊦ 2 ^ (i0 + 1) / 2 = 2 ^ i0, 0 < i0 + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i0 < 0 ⊦ 2 ^ (i0 + 1) = 0, 2 ^ (i0 + 1) = 1, 2 ^ (i0 + 1) = ~1

Rule argument

Additional information

Used simplifier rules: two-minus: ⊦ i < 0 → 2 ^ i = 0;% ⊦ i / 2 = 0 ↔ i = 0 ∨ i = 1 ∨ i = ~ 1; (ws)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; (s)