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Proof step 13 from proof twopot-mone-div (previous step)

2 ^ (j - 1) = 0, 0 < j, 0 < i, 0 ≤ j, 0 ≤ i, j ≤ i, ∀ i0, j. i→n(i0) < i→n(i) → true → j ≤ i0 ∧ 0 ≤ j → (2 ^ i0 - 1) / 2 ^ j = 2 ^ i0 / 2 ^ j - 1 ⊦ (2 ^ (i - 1) * 2) / 0 = (2 ^ (i - 1) * 2 - 1) / 0 + 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [j, 1, i1] flags:

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;%numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: j, 1, i1