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Proof step 8 from proof bitsand-zeros (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# bits ≤ 1 + # bits1 ∨ boolvar, ∀ boolvar, bits1, bits0. # bits0 < # bits → # bits0 ≤ 1 + # bits1 ∨ boolvar → bits0 = [] ∨ bitsand(bits0, boolvar ' + bits1) = bits0 ∨ true ∈ bits0 ⊦ bitsand(bits, boolvar ' + bits1) = bits, 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);