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Proof step 25 from proof and-size-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
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 = [], bits1 = bits1.first ', bitsand(bits.butlast, bits0 + bits1.butlast) = bits0 + bitsand(bits.butlast, bits1.butlast), bits.butlast.first ∧ ¬ (# bits1 - 1) + 1 < # bits

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;%n0 ≤ n ⊦ n - n0 ≤ m ↔ n ≤ m + n0;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ m < 1 ↔ m = 0;% ⊦ # bits = 0 ↔ bits = [];% ⊦ bits ≠ [] → # bits.butlast = # bits - 1;