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Proof step 5 from proof bitsor-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
bitsor(true ' + bits, bits0) = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits0 = [], bits ≠ [] ∧ # bits0 ≠ 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;