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Proof step 387 from proof bitsand-cast (previous step)

∃ bits2, bits1. bits0 = bits2 + bits1 ∧ # bits1 = 2 + # bits, sgn(mklist(boolvar, n - 1) + boolvar ' + boolvar0 ' + bits) = boolvar ' + boolvar0 ' + bits, 2 + # bits < # bits0 ⊦ lastn(n + 1 + # bits, fillsgn(n + 1 + # bits, bitsand(boolvar ' + boolvar0 ' + bits, bits0))) = lastn(n + 1 + # bits, fillsgn(n + 1 + # bits, bitsand((bits2 + mklist(boolvar, n - 1)) + boolvar ' + boolvar0 ' + bits, bits0))), n = 0, bits0 = [], boolvar ↔ boolvar0, boolvar ∧ boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 388:
# bits1 = 2 + # bits, sgn(mklist(boolvar, n - 1) + boolvar ' + boolvar0 ' + bits) = boolvar ' + boolvar0 ' + bits ⊦ bits3 = [], lastn(n + 1 + # bits, fillsgn(n + 1 + # bits, bitsand(boolvar ' + boolvar0 ' + bits, bits3 + bits1))) = lastn(n + 1 + # bits, fillsgn(n + 1 + # bits, bitsand((bits2 + mklist(boolvar, n - 1)) + boolvar ' + boolvar0 ' + bits, bits3 + bits1))), n = 0, bits3 + bits1 = [], boolvar ↔ boolvar0, boolvar ∧ boolvar0

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → n ≠ 0;% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ m = n0 → n0 < n; (forward)% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;