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Proof step 5 from proof pot2-le-pred (previous step)

2 ^ i < 2 ^ (i - 1), i ≤ 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i < 0 → j ^ i = 1 / j ^ abs(i)
Proof step 7:
i - 1 < 0, 2 ^ i < 1 / 2 ^ abs(i - 1), i ≤ 0 ⊦

Rule argument

spec: int-pot inst: name: Pot-neg seq: ⊦ i < 0 → j ^ i = 1 / j ^ abs(i) subst: [i, j] → [i - 1, 2] flags: rewrite, current

Additional information

Used simplifier rules: ls-pos-2pot-01: 0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; 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%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%sub-ls-01: ⊦ i < j - i0 ↔ i + i0 < j; from specification int-abs%ls: ⊦ i + j < j ↔ i < 0; from specification int-basic2 Used terms: i - 1, 2