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Proof step 4 from proof mod-two-ge-01 (previous step)

j < i, 0 ≤ j ⊦ 2 ^ i - 1 - (2 ^ i - 1) / 2 ^ j * 2 ^ j = 2 ^ j - 1, 2 ^ j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: two-zero: ⊦ 0 ≤ j → 2 ^ j ≠ 0;% ⊦ 1 ≤ i ↔ 0 < i; (s)%numlsf-02: ⊦ (* i) < j → i + 1 ≤ j; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j + i = i * (j + 1); (s)% ⊦ i + (j - i) = j; (s)% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i - j - i0 = i - j + i0; (s)