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Proof step 4 from proof twopot-mone-div (previous step)

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

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

Rule argument

spec: <toplevel> name: Pot-pos seq: ⊦ 0 < i → j ^ i = j ^ i -1 * j subst: [i, j] → [i, 2] flags: <all paths>

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;%eq-one: ⊦ 2 ^ i = 1 ↔ i = 0;% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i ≤ j → (j ≤ i ↔ i = j); (s)%j ≠ 0 ⊦ 0 / j = 0; (s)%j ≠ 0 ⊦ 1 / j = 1 ↔ j = 1; (s) Used terms: i, 2