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Proof step 21 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))), 0 ≤ bits2uint(lastn(m, int2bits(i))), bits2uint(lastn(m, int2bits(i))) < 2 ^ n→i(m), false ∈ int2bits(i), m ≤ # int2bits(i), lastn(m, int2bits(i)).first, 0 < i, 0 ≤ bits2uint(bits0), 0 < 2 ^ n→i(m) ⊦ bits2uint(lastn(m, int2bits(i))) = bits2int(lastn(m, int2bits(i))) + 2 ^ n→i(m), m = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ bits ≠ [] ∧ bits.first → bits2uint(bits) = bits2int(bits) + 2 ^ n→i(# bits)

Rule argument

name: ubits2bits rotate: F subst: [bits] → [lastn(m, int2bits(i))]

Additional information

Used simplifier rules: ⊦ lastn(n, bits) = [] ↔ n = 0;% ⊦ n ≤ # bits → # lastn(n, bits) = n;% ⊦ i * j + i = i * (j + 1);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i + j = i + i0 ↔ j = i0; Used terms: lastn(m, int2bits(i))