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Proof step 69 from proof cast-mod-pos-high (previous step)

j < j % 2 ^ i, 0 ≤ j, 2 ^ (i - 1) ≤ j % 2 ^ i, j < 2 ^ (i - 1), 0 ≤ 2 ^ (i - 1) + j, 0 < i, 0 < j % 2 ^ i ⊦ j % 2 ^ i = j + 2 ^ i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 70:
⊦ abs(i) < j → i % j = i

Rule argument

seq: ⊦ abs(i) < j → i % j = i subst: [j, i] → [2 ^ i, j] rotate: F spec: int-div inst: name: mod-zero-05

Additional information

Used simplifier rules: ⊦ ¬ i < i;% ⊦ 0 < j → abs(j) = j;% ⊦ 0 < j → (i ^ j = 0 ↔ i = 0);% ⊦ i = i + j ↔ j = 0;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ i + j < j ↔ i < 0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: 2 ^ i, j