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Proof step 51 from proof bitsand-fillsgn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 52:
∀ 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) + mklist(true, # bits2), false ' + bits2)) = mklist(false, n0) + bits2, n0 = 0, n0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ m + n < m + n0 ↔ n < n0;% ⊦ m < m + n ↔ n ≠ 0;% ⊦ m < n → m ≤ n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ mklist(boolvar, m + n) = mklist(boolvar, m) + mklist(boolvar, n);