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Proof step 46 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)) ⊦ fillsgn(n, bitsand(mklist(true, n), bits0)) = fillsgn(n, bits0), bits0 = [], n = 0, ¬ n < # bits0 ∧ boolvar

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
∀ 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(true, n), boolvar ' + bits2)) = fillsgn(n, boolvar ' + bits2), ¬ n ≤ # bits2 ∧ boolvar

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ boolvar ' + bits ≠ [];% ⊦ boolvar + bits = boolvar ' + bits;