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Proof step 60 from proof bitsand-fillsgn (previous step)

# mklist(true, # bits2) = # bits2 ∧ mklist(true, n0) ≠ [] ∧ false ' ≠ [] ∧ mklist(true, # bits2) ≠ [], ∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n0 + # bits2, bitsand(mklist(true, n0), false ') + bitsand(mklist(true, # bits2), bits2)) = mklist(false, n0) + bits2, n0 = 0, n0 = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 61:
∀ n, bits, bits1. # bits1 < #(false + bits2) → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(n0 + # bits2, false ' + bitsand(mklist(true, # bits2), bits2)) = mklist(false, n0) + bits2, n0 = 0, n0 = 1, bits2 = []

Rule argument

Additional information

Used simplifier rules: and-zero-03: ⊦ bits ≠ [] → bitsand(bits, false ') = false ';% ⊦ boolvar ' ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ mklist(boolvar, m) = [] ↔ m = 0;% ⊦ # mklist(boolvar, n) = n;