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Proof step 17 from proof add-ones-grlen (previous step)

bits = boolvar + bits2, # bits = n + 1 + # bits1, # bits1 < # bits, ∀ bits0, bits, n. # bits0 < # bits1 → # bits = n + 1 + # bits0 ∧ # bits0 < # bits → bits = [] ∨ bitsand(bits, true ' + bits0) = (true ' + mklist(true, n)) + bits0 ∨ false ∈ bits ⊦ bits = boolvar ', bits1 = [], bits = [], bitsand(bits.butlast, true ' + bits1.butlast) + (bits1.last ∧ bits.last)' = (true ' + mklist(true, n)) + bits1, false ∈ bits

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# bits2 = n + # bits1, # bits1 ≤ # bits2, ∀ bits0, bits, n. # bits0 < # bits1 → # bits = n + 1 + # bits0 ∧ # bits0 < # bits → bits = [] ∨ bitsand(bits, true ' + bits0) = (true ' + mklist(true, n)) + bits0 ∨ false ∈ bits ⊦ bitsand(true ' + bits2.butlast, true ' + bits1.butlast) + (bits2.last ∧ bits1.last)' = (true ' + mklist(true, n)) + bits1, bits1 = [], bits2 = [], false ∈ bits2

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ boolvar ∈ boolvar0 ' + bits ↔ ¬(boolvar ↔ boolvar0) ∧ ¬ boolvar ∈ bits;% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m < n + 1 ↔ ¬ n < m;% ⊦ bits ≠ [] → ((boolvar ' + bits).last ↔ bits.last);% ⊦ bits ≠ [] → (bits0 + bits).butlast = bits0 + bits.butlast;% ⊦ boolvar ' + bits ≠ [];% ⊦ bits = bits + bits0 ↔ bits0 = [];% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ boolvar + bits = boolvar ' + bits;