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

0 < i, 0 < i, 1 < i ⊦ (2 ^ i -1 * 2 - 1) % 2 = 1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: mod-two-03 seq: ⊦ 0 < i → (i * 2 - 1) % 2 = 1 subst: [i] → [2 ^ i -1] flags: rewrite

Additional information

Used simplifier rules: pot2-pos: ⊦ 0 < 2 ^ i ↔ 0 ≤ i;%sub-pred: ⊦ i -1 = i - 1; from specification int-abs%sub-le-01: ⊦ i ≤ j - i0 ↔ i + i0 ≤ j; 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%leone: ⊦ 1 ≤ i ↔ 0 < i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used terms: 2 ^ i -1