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Proof step 31 from proof bitsand-fillsgn (previous step)

¬ true ∈ mklist(false, n) ∧ # bits0 < # mklist(false, n) ∧ ¬ bits0.first ∧ mklist(false, n) ≠ [] ∧ bits0 ≠ [], ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], fillsgn(n, mklist(false, # bits0)) = mklist(false, n), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
# bits0 < n, ∀ n, bits, bits1. # bits1 < # bits0 → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits0 = [], fillsgn(n, mklist(false, # bits0)) = mklist(false, n), bits0.first

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ true ∈ mklist(false, n);% ⊦ m < n → n ≠ 0;% ⊦ # mklist(boolvar, n) = n;% ⊦ mklist(boolvar, m) = [] ↔ m = 0;