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Proof step 185 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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