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Proof step 20 from proof bitsand-cast (previous step)

n0 ≠ 0, 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 ⊦ lastn(n, fillsgn(n, bitsand(boolvar ' + boolvar0 ' + bits1, bits0))) = lastn(n, fillsgn(n, bitsand((mklist(boolvar, n0 -1) + boolvar ') + boolvar0 ' + bits1, bits0))), bits0 = [], n0 = 0, boolvar ↔ boolvar0, ¬ false ∈ mklist(boolvar, n0 -1) + boolvar ' ∧ boolvar0 ∧ ¬ false ∈ bits1, ¬ true ∈ mklist(boolvar, n0 -1) + boolvar ' ∧ ¬ boolvar0 ∧ ¬ true ∈ bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits;% ⊦ m < n ∧ n = n0 → m < n0; (forward)% ⊦ bits = bits0 + boolvar ' + bits1 → boolvar ∈ bits; (forward)% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ ¬ true ∈ mklist(false, n);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ ¬ false ∈ mklist(true, n);% ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ n -1 = n - 1;