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Proof step 2 from proof int2int-lastn-2pot (previous step)

i % 2 ^ n→i(m) = 0, m ≤ # int2bits(i) ⊦ m = 0, bits2int(lastn(m, int2bits(i))) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)
Proof step 4:
2 ^ n→i(m) ≠ 0, ∃ j0. i = j0 * 2 ^ n→i(m), m ≤ # int2bits(i) ⊦ m = 0, bits2int(lastn(m, int2bits(i))) = 0

Rule argument

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

Additional information

Used simplifier rules: ⊦ 2 ^ n→i(m) ≠ 0; Used terms: 2 ^ n→i(m), i