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Proof step 40 from proof getShort-isi (previous step)

0 ≤ b2i(byi), 0 ≤ b2i(byi) ⊦ s2i(i2s(bor(256 * bits2uint(uint2bits(b2i(byi))), band(b2i(byj), 255)))) = bor(256 * bits2uint(uint2bits(b2i(byi))), band(b2i(byj), 255)), ¬ bor(256 * bits2uint(uint2bits(b2i(byi))), band(b2i(byj), 255)) < ~32768 ∧ ¬ 32767 < bor(256 * bits2uint(uint2bits(b2i(byi))), band(b2i(byj), 255))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
0 ≤ b2i(byi) ⊦ s2i(i2s(bor(256 * b2i(byi), band(b2i(byj), 255)))) = bor(256 * b2i(byi), band(b2i(byj), 255)), ¬ bor(256 * b2i(byi), band(b2i(byj), 255)) < ~32768 ∧ ¬ 32767 < bor(256 * b2i(byi), band(b2i(byj), 255))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i * j = j * i;% ⊦ 0 ≤ i → bits2uint(uint2bits(i)) = i;