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Proof step 19 from proof bitsand-first-samesize (previous step)

bits1 ≠ [] ∧ bits = true ', # bits1 = 1, # bits = 1, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < 1 → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bits1 = [], bits = [], bitsand(boolvar ' + bits1, boolvar0 ' + bits) = (boolvar ∧ boolvar0)' + bits1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
# bits1 = 1, ∀ boolvar0, boolvar, bits1, bits0. # bits0 < 1 → # bits1 = # bits0 → bits0 = [] ∨ bitsand(boolvar ' + bits1, boolvar0 ' + bits0) = (boolvar ∧ boolvar0)' + bitsand(bits1, bits0) ⊦ bitsand(boolvar ' + bits1, boolvar0 ' + true ') = (boolvar ∧ boolvar0)' + bits1, bits1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ' ≠ [];% ⊦ # boolvar ' = 0 +1;