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Proof step 2 from proof mod-two-ls-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div name: mod-zero-03 seq: 0 < j, 0 ≤ i, i < j ⊦ i % j = i subst: [j, i] → [2 ^ j, 2 ^ i - 1] flags: <all paths>

Additional information

Used simplifier rules: two-le: ⊦ 0 ≤ i → (2 ^ i ≤ 2 ^ j ↔ i ≤ j);%two-lez: ⊦ 0 ≤ i → ¬ 2 ^ i ≤ 0;% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 1 ≤ i ↔ 0 < i; (s)% ⊦ i - j < i0 ↔ i < j + i0; (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)%numlef-01: ⊦ (* i) ≤ j → i - 1 < j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ ~1 < i ↔ 0 ≤ i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss) Used terms: 2 ^ j, 2 ^ i - 1