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Proof step 15 from proof add-ones-grlen (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
# 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 = bits.first ', bits1 = [], bits = [], bitsand(bits.butlast, true ' + bits1.butlast) + (bits1.last ∧ bits.last)' = (true ' + mklist(true, n)) + bits1, false ∈ bits

Rule argument

Additional information

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