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Proof step 9 from proof bitsor-bi-ones (previous step)

¬(¬ false ∈ bits ∧ (true ' + bits2).first ∧ # bits ≤ #(true ' + bits2) ∧ bits ≠ [] ∧ true ' + bits2 ≠ []) ⊦ bits = [], bits2int(bitsor(bits, true ' + bits2) + bits1) = bits2int(true ' + bits1), false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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