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Proof step 10 from proof and-notone-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
bitsand(bits.butlast, false ' + bits0.butlast) + (bits0.last ∧ bits.last)' = true ' + bits1, ∀ bits2, bits0, bits1. # bits2 < # bits → bitsand(bits2, false ' + bits0) = true ' + bits1 → bits2 = [] ⊦ bits0 = [], bits = bits.first ', bits = [], (¬ bits0.last ∨ ¬ bits.last) ∧ ¬ (bits0.last ∧ bits.last) ∈ bits1, ¬ true ∈ bitsand(bits.butlast, false ' + bits0.butlast) ∧ (¬ bits0.last ∨ ¬ bits.last)

Rule argument

Additional information

Used simplifier rules: ⊦ boolvar ∈ boolvar0 ' ↔ boolvar ↔ boolvar0;% ⊦ boolvar ∈ bits + bits0 ↔ ¬ ¬(boolvar ∈ bits ∨ boolvar ∈ bits0);% ⊦ boolvar ' + bits = bits0 → boolvar ∈ bits0; (forward)% ⊦ bits + boolvar ' = bits0 → boolvar ∈ bits0; (forward)% ⊦ 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 ';