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Proof step 80 from proof byte-law-and-one (previous step)

¬ # bitsand(fillsgn(8, int2bits(i)), int2bits(j)) ≤ 8, 8 = # fillsgn(8, int2bits(i)), 8 = # fillsgn(8, bitsand(int2bits(i), int2bits(j))), # int2bits(i) < 8, # int2bits(j) < 8 ⊦ 8 = # fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j))), fillsgn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), bitsand(fillsgn(8, int2bits(i)), int2bits(j)) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ n < # bits → lastn(n, fillsgn(n, bits)) = lastn(n, bits);% ⊦ ¬ m ≤ n ↔ n < m;