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Proof step 136 from proof byte-law-and-one (previous step)

int2bits(i) ≠ [] ∧ bits ≠ [], # bits = 8, int2bits(j) = bits0 + bits, 7 = # fillsgn(7, int2bits(i)), 8 = # fillsgn(8, bitsand(bits, int2bits(i))), 8 = # fillsgn(8, bitsand(int2bits(i), bits)), # int2bits(i) < 8, 0 ≤ i ⊦ bits = [], fillsgn(8, int2bits(i)) = [], bits0 + bits = [], fillsgn(8, bitsand(bits, int2bits(i))) = fillsgn(8, bitsand(int2bits(i), bits))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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