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Proof step 41 from proof shiftright-neg-div (previous step)

2 ^ (abs(j) - 2) = 0, i0 >> (abs(j) - 1) = (1 + i0) / 0, i0 * 2 + j1 < 0, 1 < abs(j), i0 * 2 + j1 < ~1, j1 ≤ 0, i0 ≤ 0, abs(j1) < 2, ∀ i, j0. i→n(j0) < i→n(j) → i + 2 ^ (abs(j0) - 1) < 0 → j0 = 0 ∨ i >> abs(j0) = (i + 1 - 2 ^ abs(j0)) / 2 ^ abs(j0) ⊦ j = 0, abs(j) = 1, (1 + i0) / 0 = (2 * (i0 + 1) + j1) / 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 42:
⊦ 0 ≤ j → 2 ^ j ≠ 0

Rule argument

seq: ⊦ 0 ≤ j → 2 ^ j ≠ 0 subst: [j] → [abs(j) - 2] rotate: F spec: int-pot inst: name: two-zero

Additional information

Used simplifier rules: ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ ¬ i ≤ j ↔ j < i;%i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; Used terms: abs(j) - 2