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Proof step 170 from proof byte-law-and-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n < m0 → m + n < m0 + n0;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ m +1 = m + 1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m;% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;