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Proof step 28 from proof bitsand-fillsgn (previous step)

bits0 = boolvar + bits2, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], fillsgn(n, bitsand(mklist(false, n), bits0)) = mklist(false, n), n = 0, ¬ n ≤ # bits0 ∧ ¬ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
∀ n, bits, bits1. # bits1 < #(boolvar + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n = 0, fillsgn(n, bitsand(mklist(false, n), boolvar ' + bits2)) = mklist(false, n), ¬ n ≤ 1 + # bits2 ∧ ¬ boolvar

Rule argument

Additional information

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