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Proof step 92 from proof bitsand-cast (previous step)

boolvar ' + boolvar0 ' + bits1 ≠ [] ∧ bits2 + bits ≠ [], # bits = 2 + # bits1, sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), n0 + 1 + # bits1 < n ⊦ bits2 = [], lastn(n, fillsgn(n, bitsand(bits2 + bits, boolvar ' + boolvar0 ' + bits1))) = lastn(n, fillsgn(n, bitsand(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1, bits2 + bits))), n0 = 0, boolvar ↔ boolvar0, bits2 + bits = [], boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 93:
# bits = 2 + # bits1, sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), n0 + 1 + # bits1 < n ⊦ bits2 = [], lastn(n, fillsgn(n, bitsand(bits2 + bits, boolvar ' + boolvar0 ' + bits1))) = lastn(n, fillsgn(n, bitsand(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1, bits2 + bits))), n0 = 0, boolvar ↔ boolvar0, bits2 + bits = [], boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];