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Proof step 6 from proof two-minus (previous step)

i < 0, 2 ^ abs(i) ≤ 1 ⊦ 1 / 2 ^ abs(i) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ j < 0 → abs(j) = 0 - j
Proof step 8:
i < 0, 2 ^ (0 - i) ≤ 1 ⊦ 1 / 2 ^ (0 - i) = 0

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i] flags: <all paths>

Additional information

Used terms: i