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Proof step 26 from proof int2int-lastn-negpos (previous step)

bits2int(bits) * 2 ^ n→i(m) + bits2int(lastn(m, int2bits(i))) = i, int2bits(i) = bits + lastn(m, int2bits(i)), # bits = # int2bits(i) - m, bits2int(lastn(m, int2bits(i))) % 2 ^ n→i(m) = 0, m ≤ # int2bits(i), i < 0, bits.first ⊦ bits = [], m = 0, i % 2 ^ n→i(m) = 0, bits2int(lastn(m, int2bits(i))) = i % 2 ^ n→i(m) + 2 ^ n→i(m), lastn(m, int2bits(i)).first

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j)

Rule argument

seq: j ≠ 0 ⊦ i % j = 0 ↔ (∃ j0. i = j0 * j) subst: [j, i] → [2 ^ n→i(m), bits2int(lastn(m, int2bits(i)))] rotate: F spec: int-div inst: name: modex

Additional information

Used simplifier rules: ⊦ 2 ^ n→i(m) ≠ 0;% ⊦ i * j = j * i;% ⊦ i < j ∧ i = i0 → i0 < j; (forward)% ⊦ i * j + i * j0 = i * (j + j0);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;%j ≠ 0 ⊦ (i * j) % j = 0;% ⊦ i + 0 = i;% ⊦ i * j = i ↔ i = 0 ∨ j = 1; Used terms: 2 ^ n→i(m), bits2int(lastn(m, int2bits(i)))