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Proof step 57 from proof bitsand-fillsgn (previous step)

mklist(true, n0) ≠ [] ∧ false ' = false ', ∀ n, bits, bits1. # bits1 < #(false + []) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n0 = 1, n0 = 0, fillsgn(n0, false ') = mklist(false, n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
∀ n, bits, bits1. # bits1 < #(false + []) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ n0 = 1, n0 = 0, fillsgn(n0, false ') = mklist(false, n0)

Rule argument

Additional information

Used simplifier rules: ⊦ mklist(boolvar, m) = [] ↔ m = 0;