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Proof step 20 from proof getShort-isi (previous step)

¬(# mklist(false, 8) ≠ 0 ∧ # int2bits(b2i(byi)) ≠ 0 ∧ ¬ int2bits(b2i(byi)).first) ⊦ s2i(i2s(bor(bits2int(int2bits(b2i(byi)) + mklist(false, 8)), band(b2i(byj), 255)))) = bor(bits2int(int2bits(b2i(byi)) + mklist(false, 8)), band(b2i(byj), 255)), ¬ bor(bits2int(int2bits(b2i(byi)) + mklist(false, 8)), band(b2i(byj), 255)) < ~32768 ∧ ¬ 32767 < bor(bits2int(int2bits(b2i(byi)) + mklist(false, 8)), band(b2i(byj), 255))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ int2bits(j).first ↔ j < 0;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ # mklist(boolvar, n) = n;% ⊦ ¬ i < j ↔ j ≤ i;