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Proof step 18 from proof and-size-one (previous step)

bits = boolvar + bits3, bits1 = bits1.first ', boolvar, # 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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bits1 = bits1.first ', # bits3 < # bits1, ∀ bits2, bits1, bits0. # bits2 < #(true + bits3) → # bits2 ≤ # bits1 ∧ bits2.first → bits2 = [] ∨ bitsand(bits2, bits0 + bits1) = bits0 + bitsand(bits2, bits1) ⊦ bitsand(true ' + bits3.butlast, bits0 + bits1.butlast) + (bits1.first ∧ bits3.last)' = bits0 + bitsand(true ' + bits3, bits1), bits1 = [], # bits0 + # bits1 = 0, bits3 = [], bits0 = []

Rule argument

Additional information

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