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Proof step 27 from proof bitsand-cast (previous step)

# bits0 ≤ #(boolvar ' + boolvar0 ' + bits1) ∧ bits0 ≠ [] ∧ bits0.first, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, # bits0 ≤ 2 + # bits1, n0 + 1 + # bits1 < n, bits0.first ⊦ bits0 = [], boolvar ↔ boolvar0, n0 = 0, lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits0))) = lastn(n, fillsgn(n, mklist(boolvar, n0 - 1) + bitsand(boolvar ' + boolvar0 ' + bits1, bits0))), boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n0 + 1 + # bits1 < n, bits0.first, # bits0 ≤ 2 + # bits1 ⊦ boolvar ↔ boolvar0, n0 = 0, lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits0))) = lastn(n, fillsgn(n, mklist(boolvar, n0 - 1) + bitsand(boolvar ' + boolvar0 ' + bits1, bits0))), bits0 = [], boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;