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Proof step 262 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 263:
int2bits(j) = bits2 + bits1, # bits1 = 8, int2bits(i) = bits0 + bits, # bits = 8, # bitsand(bits, bits1) < 8, 8 ≤ # bitsand(bits0, bits2) + # bitsand(bits, bits1) ⊦ lastn(8, bitsand(bits0, bits2) + bitsand(bits, bits1)) = lastn(8, fillsgn(8, bitsand(bits2 + bits1, bits))), bits2 + bits1 = [], bits = [], bits2 = [], bits0 = []

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;