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Proof step 174 from proof byte-law-and-one (previous step)

7 = # fillsgn(7, int2bits(i)), # bits1 = # int2bits(i), int2bits(j) = (bits0 + bits2) + bits1, # bits2 + # bits1 = 8, # bits1 < 8, # bits1 < # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), # bits1 < # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), i < 0, 8 ≤ # bits0 + # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bitsand(bits2 + bits1, fillsgn(8, int2bits(i))), 8 ≤ # bits0 + # bits2 + # bitsand(bits1, int2bits(i)), # bits1 < # bits0 + # bits2 + # bitsand(bits1, int2bits(i)) ⊦ bits2 = [], bitsand(bits2 + bits1, fillsgn(8, int2bits(i))) = [], lastn(8, (bits0 + bits2) + bitsand(bits1, int2bits(i))) = lastn(8, bitsand(bits2 + bits1, fillsgn(8, int2bits(i))))

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(i), 8] rotate: F spec: bitopsaux inst: name: fill-fill

Additional information

Used simplifier rules: ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ int2bits(j) ≠ [];% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) Used terms: int2bits(i), 8