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Proof step 16 from proof and-size-one (previous step)

¬(1 < # bits ∧ 1 < # bits1), bits.first, # bits ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', # bits0 + # bits1 = 0, bits1 = [], bits = [], bitsand(bits.butlast, bits0 + bits1.butlast) + (bits1.last ∧ bits.last)' = bits0 + bitsand(bits, bits1), bits0 = [] ∧ bits1 = bits1.first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
bits1 = bits1.first ', bits.first, # bits ≤ # bits1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', # bits0 + # bits1 = 0, bits1 = [], bits = [], bitsand(bits.butlast, bits0 + bits1.butlast) + (bits.last ∧ bits1.first)' = bits0 + bitsand(bits, bits1), bits0 = []

Rule argument

Additional information

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