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Proof step 8 from proof bitsor-zeroes-true (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ bits1, bits0. # bits0 < # bits → true → bits0 = [] ∨ bitsor(bits0, true ' + bits1) = true ' + bits1 ∨ true ∈ bits0 ⊦ bitsor(bits, true ' + bits1) = true ' + bits1, bits = [], true ∈ bits, # bits ≠ 1 ∧ bits1 ≠ []

Rule argument

Additional information

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