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Proof step 20 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.butlast, bits1.butlast) + (bits.last ∧ bits1.last)), bits0 = [] ∧ bits1 = bits1.first '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
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.butlast, bits1.butlast) + (bits.last ∧ bits1.last)', bits1 = bits1.first '

Rule argument

Additional information

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