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Proof step 19 from proof cast-law-add-one (previous step)

j % 2 ^ i = 0, 0 < i, j < 0 ⊦ cast(i, j + j0) = cast(i, cast(i, j) + j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 21:
2 ^ i ≠ 0, ∃ j0. j = j0 * 2 ^ i, 0 < i, j < 0 ⊦ cast(i, j + j0) = cast(i, cast(i, j) + j0)

Rule argument

seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [2 ^ i, j] rotate: F spec: int-div inst: name: modex

Additional information

Used simplifier rules: ⊦ 0 < j → (i ^ j = 0 ↔ i = 0); Used terms: 2 ^ i, j