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Proof step 4 from proof div2-2pot-03 (previous step)

0 ≤ j, 0 ≤ i0 ⊦ (j * 2 ^ (i0 + 1) + 1) / 2 = j * 2 ^ i0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ 0 < i → j ^ i = j ^ i -1 * j
Proof step 6:
0 < i0 + 1, 0 ≤ j, 0 ≤ i0 ⊦ (j * 2 ^ (i0 + 1)-1 * 2 + 1) / 2 = j * 2 ^ i0

Rule argument

spec: int-pot inst: name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [i0 + 1, 2] flags: rewrite, current

Additional information

Used simplifier rules: numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%mone-06: ⊦ ~1 < i ↔ 0 ≤ i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%lels-04: ⊦ i ≤ j → (¬ ¬ j < i ↔ false); 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 Used terms: i0 + 1, 2