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Proof step 5 from proof bitsand-zeros (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
# 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

Proof step 5 from proof bitsand-zeros (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information