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Proof step 30 from proof bitsand-cast (previous step)

¬ n ≤ # bitsand(boolvar ' + boolvar0 ' + bits1, bits0), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
sgn(mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1) = boolvar ' + boolvar0 ' + bits1, n = # fillsgn(n, mklist(boolvar, n0 - 1) + boolvar ' + boolvar0 ' + bits1), # bitsand(boolvar ' + boolvar0 ' + bits1, bits0) < n, # bits0 ≤ 2 + # bits1, bits0.first, n0 + 1 + # bits1 < n ⊦ bits0 = [], lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits0))) = lastn(n, fillsgn(n, mklist(boolvar, n0 - 1) + bitsand(boolvar ' + boolvar0 ' + bits1, bits0))), n0 = 0, boolvar ↔ boolvar0, boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;