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Proof step 468 from proof bitsand-cast (previous step)

∃ bits0, bits1. bits3 = bits0 + bits1 ∧ # bits1 = m, sgn(mklist(true, m) + true ' + false ' + bits) = true ' + false ' + bits, # bits1 = 2 + # bits, m < # bits3 ⊦ bits3 + bits1 = [], bits1 = [], lastn(2 + # bits + m, fillsgn(2 + # bits + m, bits3 + bitsand(bits1, true ' + false ' + bits))) = lastn(2 + # bits + m, fillsgn(2 + # bits + m, bitsand(bits2 + mklist(true, m), bits3) + bitsand(bits1, true ' + false ' + bits))), bits2 + mklist(true, m) = [], bits3 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 469:
sgn(mklist(true, # bits0) + true ' + false ' + bits) = true ' + false ' + bits, # bits1 = 2 + # bits ⊦ (bits4 + bits0) + bits1 = [], bits1 = [], lastn(2 + # bits + # bits0, fillsgn(2 + # bits + # bits0, (bits4 + bits0) + bitsand(bits1, true ' + false ' + bits))) = lastn (2 + # bits + # bits0, fillsgn(2 + # bits + # bits0, bitsand(bits2 + mklist(true, # bits0), bits4 + bits0) + bitsand(bits1, true ' + false ' + bits))), bits2 + mklist(true, # bits0) = [], bits4 + bits0 = [], bits4 = []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits;% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < m + n ↔ n ≠ 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ #(bits + bits0) = # bits + # bits0;