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Proof step 3 from proof two-minus (previous step)

i < 0 ⊦ 1 / 2 ^ abs(i) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
0 < j, 0 ≤ i, i < j ⊦ i / j = 0
Proof step 5:
i < 0 ⊦ 1 / 2 ^ abs(i) = 0, 0 < 2 ^ abs(i) ∧ 0 ≤ 1 ∧ 1 < 2 ^ abs(i)

Rule argument

spec: int-div name: div-zero seq: 0 < j, 0 ≤ i, i < j ⊦ i / j = 0 subst: [j, i] → [2 ^ abs(i), 1] flags: <all paths>

Additional information

Used terms: 2 ^ abs(i), 1