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Proof step 224 from proof int2bytes-rec-neg-05 (previous step)

# int2bits(j % 2 ^ (i * 8)) < 8 * i→n(i), j + 2 ^ (i * 8 - 1) < 0, j + 2 ^ (i * 8) ≤ 0, j + 2 ^ (i * 8) < i, j % 2 ^ (i * 8) < 0, 0 < i, # int2bits(j % 2 ^ (i * 8)) ≤ 8 * i→n(i) + 8 ⊦ i = 0, 1 + 8 * i→n(i) = # int2bits(j % 2 ^ (i * 8)), mklist(true, 8 * i→n(i) - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8)) = [], j % 2 ^ (i * 8) = 0, bits2bytes(mklist(true, 8 * i→n(i) - # int2bits(j % 2 ^ (i * 8))) + int2bits(j % 2 ^ (i * 8))) = bits2bytes(fillsgn(8 * i→n(i) + 8, int2bits(j % 2 ^ (i * 8)))).rest

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ # bits ≤ n ∧ bits ≠ [] → fillsgn(n, bits) = mklist(bits.first, n - # bits) + bits subst: [bits, n] → [int2bits(j % 2 ^ (i * 8)), 8 * i→n(i) + 8] rotate: F spec: bitopsaux inst: name: fill-fill

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ m * n = n * m;% ⊦ m + n = n + m;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m ≤ n → (n < m ↔ false);% ⊦ int2bits(j) ≠ []; Used terms: int2bits(j % 2 ^ (i * 8)), 8 * i→n(i) + 8