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Proof step 112 from proof byte-law-and-one (previous step)

¬ # bitsand(bits, int2bits(i)) ≤ 8, # bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < 8, 0 ≤ i ⊦ 8 = # fillsgn(8, bitsand(bits, int2bits(i))), bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits, int2bits(i)))) = lastn(8, fillsgn(8, bitsand(bits, false ' + fillsgn(7, int2bits(i)))))

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;