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Proof step 6 from proof bitsand-app-zeros-01 (previous step)

¬(1 < #(false ' + bits0) ∧ 1 < # bits1), ∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ sgn(bitsand(bits + false ' + bits0, bits2)) = sgn(bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ sgn(bitsand(bits + false ' + bits0, bits1)) = sgn(bitsand(false ' + bits0, bits1)), bits1 = [], true ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ sgn(bitsand(bits + false ' + bits0, bits2)) = sgn(bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ sgn(bitsand(bits + false ' + bits0, bits1)) = sgn(bitsand(false ' + bits0, bits1)), bits1 = [], true ∈ bits, bits0 ≠ [] ∧ # bits1 ≠ 1

Rule argument

Additional information

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