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Proof step 10 from proof and-size-one (previous step)

bits = true ', bits.first, # bits ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits1 = [], bitsand(bits, bits0 + bits1) = bits0 + bitsand(bits, bits1), bits = [], # bits0 + # bits1 ≠ 1 ∧ bits ≠ true '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
1 ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # true ' → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits1 = [], bitsand(true ', bits0 + bits1) = bits0 + bits1

Rule argument

Additional information

Used simplifier rules: and-onel-01: ⊦ bits ≠ [] → bitsand(true ', bits) = bits;% ⊦ # boolvar ' = 0 +1;% ⊦ boolvar '.first ↔ boolvar;% ⊦ boolvar ' ≠ [];