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Proof step 27 from proof and-size-one (previous step)

bits1 = bits3 + boolvar ', 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 + bits3) = bits0 + bitsand(bits.butlast, bits3), bits.butlast.first ∧ ¬ (# bits1 - 1) + 1 < # bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
bits.first, # bits ≤ # bits3 + 1, ∀ bits2, bits1, bits0. # bits2 < # bits → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bits = true ', bits = [], bits3 + boolvar ' = (bits3 + boolvar ').first ', bitsand(bits.butlast, bits0 + bits3) = bits0 + bitsand(bits.butlast, bits3), bits.butlast.first

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n ∧ n ≤ n0 → m < n0; (forward)% ⊦ m ≤ n ∧ n < n0 → m < n0; (forward)% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ 0 + m = m;% ⊦ n ≤ m → m + n0 - n = (m - n) + n0;% ⊦ bits + boolvar ' ≠ [];% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ m + n = n + m;