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Proof step 159 from proof byte-law-and-one (previous step)

∃ bits0, bits1. bits = bits0 + bits1 ∧ # bits1 = # int2bits(i), int2bits(j) = bits0 + bits, # bits = 8, 7 = # fillsgn(7, int2bits(i)), # int2bits(i) < # bitsand(bits, fillsgn(8, int2bits(i))), i < 0, # int2bits(i) < 8, 8 ≤ # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))), # int2bits(i) < # bits0 + # bitsand(bits, fillsgn(8, int2bits(i))), 8 ≤ # bitsand(bits, fillsgn(8, int2bits(i))) ⊦ bits0 + bits = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(i)))) = lastn(8, bitsand(bits, true ' + fillsgn(7, int2bits(i)))), bitsand(bits, true ' + fillsgn(7, int2bits(i))) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < m + n ↔ n ≠ 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ int2bits(j) ≠ [];% ⊦ m < n → n ≠ 0;% ⊦ m + n = n + m;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)%0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;