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Proof step 114 from proof bytes2int-int2bytes-len-eq (previous step)

~256 < bytes2int(bytes1) * 256 + band(b2i(by), 255) ∧ bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~128, # bytes1 + 1 = #(i2b(~1) ' + i2b(bytes2int(bytes1) * 256 + band(b2i(by), 255)) '), bytes2int(bytes1) * 256 + band(b2i(by), 255) < 0, bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~128, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 = [], bytes1 + by ' = i2b(~1) ' + i2b(bytes2int(bytes1) * 256 + band(b2i(by), 255)) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bytes + bytes0 = bytes1 + bytes0 ↔ bytes = bytes1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%n0 < n ⊦ (* n) = m0 + (* n0) ↔ n - n0 = m0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ i2b(band(255, b2i(by))) = by;% ⊦ i2b(i * 256 + j) = i2b(j);