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Proof step 243 from proof byte-law-and-one (previous step)

int2bits(i) = bits0 + bits, # bits = 8, # int2bits(j) ≤ # bitsand(bits, int2bits(j)), j < 0, j < ~64, ~128 ≤ j, # int2bits(j) ≤ 8, 8 ≤ # bitsand(bits, int2bits(j)) ⊦ bits0 = [], bitsand(bits, int2bits(j)) = [], lastn(8, fillsgn(8, bits0 + bitsand(bits, int2bits(j)))) = lastn(8, bitsand(bits, int2bits(j)))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 244:
⊦ n ≤ # bits → fillsgn(n, bits) = bits

Rule argument

seq: ⊦ n ≤ # bits → fillsgn(n, bits) = bits subst: [bits, n] → [bits0 + bitsand(bits, int2bits(j)), 8] rotate: F spec: bitopsaux inst: name: fillsgn-longer

Additional information

Used simplifier rules: ⊦ n ≤ # bits → lastn(n, bits0 + bits) = lastn(n, bits);% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m + n < m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m; Used terms: bits0 + bitsand(bits, int2bits(j)), 8