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Proof step 84 from proof bitsand-fillsgn (previous step)

m + 1 ≠ 0 ∧ bitsand(fillsgn(m, bits2), bits3) ≠ [], ∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ bits3 = [], bits2 = [], fillsgn((m + 1)-1, bitsand(fillsgn(m, bits2), bits3)) + (boolvar ∧ boolvar0) = fillsgn(m + 1, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)'), fillsgn(m, bits2) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 85:
∀ n, bits, bits1. # bits1 < #(bits3 + boolvar0 ') → true → bits = [] ∨ bits1 = [] ∨ fillsgn(n, bitsand(fillsgn(n, bits), bits1)) = fillsgn(n, bitsand(bits, bits1)) ⊦ fillsgn(m, bits2) = [], bits3 = [], bits2 = [], fillsgn(m, bitsand(fillsgn(m, bits2), bits3)) + (boolvar ∧ boolvar0)' = fillsgn(m + 1, bitsand(bits2, bits3) + (boolvar ∧ boolvar0)')

Rule argument

Additional information

Used simplifier rules: and-notempty: ⊦ bits ≠ [] ∧ bits0 ≠ [] → bitsand(bits, bits0) ≠ [];% ⊦ m + 1 ≠ 0;% ⊦ bits + boolvar = bits + boolvar ';% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ (m + 1)-1 = m;