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Proof step 15 from proof int2int-lastn-posneg (previous step)

1 + # bits0 = # int2bits(i) - m, int2bits(i) = (false ' + bits0) + lastn(m, int2bits(i)), i = bits2uint(bits0) * 2 ^ n→i(m) + bits2uint(lastn(m, int2bits(i))), false ∈ int2bits(i), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i ⊦ (bits2uint(bits0) * 2 ^ n→i(m) + bits2uint(lastn(m, int2bits(i)))) % 2 ^ n→i(m) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: 0 ≤ i0 * j + j0, 0 ≤ j0, j0 < j ⊦ (i0 * j + j0) % j = j0 subst: [j0, j, i0] → [bits2uint(lastn(m, int2bits(i))), 2 ^ n→i(m), bits2uint(bits0)] rotate: F spec: int-div inst: name: modeq-pos

Additional information

Used terms: bits2uint(lastn(m, int2bits(i))), 2 ^ n→i(m), bits2uint(bits0)