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Proof step 179 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0;% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ n < m → m - n ≠ 0;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;