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Proof step 100 from proof byte-law-and-one (previous step)

int2bits(i) ≠ [] ∧ bits0 + bits ≠ [], 8 = # fillsgn(8, int2bits(i)), # bits = 8, int2bits(j) = bits0 + bits, # int2bits(i) < 8, 0 ≤ i ⊦ fillsgn(8, int2bits(i)) = [], lastn(8, fillsgn(8, bitsand(bits0 + bits, int2bits(i)))) = lastn(8, fillsgn(8, bitsand(bits, fillsgn(8, int2bits(i))))), bits0 + bits = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j) ≠ [];% ⊦ boolvar ' + bits ≠ [];% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ int2bits(j).first ↔ j < 0;%⊦ bits ≠ [] ∧ # bits < n → fillsgn(n, bits) = bits.first ' + fillsgn(n -1, bits);