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Proof step 60 from proof byte-law-and-one (previous step)

# int2bits(i) < # int2bits(j), 8 = # fillsgn(8, int2bits(i)), # int2bits(i) < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < 8, # int2bits(i) < 8, 8 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) + 1 < # bitsand(int2bits(i), int2bits(j)), # int2bits(j) < # bitsand(int2bits(i), int2bits(j)), # int2bits(i) + 1 < # bitsand(int2bits(i), int2bits(j)) ⊦ lastn(8, bitsand(int2bits(i), int2bits(j))) = lastn(8, fillsgn(8, bitsand(fillsgn(8, int2bits(i)), int2bits(j)))), 8 = # fillsgn(8, bitsand(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 + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n < n0 ⊦ m + (* n) < (* n0) ↔ m < n0 - n;% ⊦ m +1 = m + 1;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ n < n0 ∧ n0 < n1 → n +1 < n1; (forward)