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Proof step 18 from proof add-ones-grlen-02 (previous step)

bits0 = boolvar0 + bits3, bits0.butlast = boolvar ' + bits2, boolvar ∈ bits0.butlast, # bits0 ≤ # bits, boolvar ∈ bits0, boolvar ∨ bitsand(bits.butlast, bits0.butlast) = bits0.butlast, ∀ bits, bits1. # bits1 < # bits0 → # bits1 ≤ # bits → bits1 = [] ∨ bitsand(bits, bits1) = bits1 ∨ false ∈ bits ∨ bits1.first ⊦ bits0 = false ', # bits0 = 1, bits = [], # bits = 1, bitsand(bits.butlast, bits0.butlast) + (bits.last ∧ bits0.last)' = bits0, bits0 = [], false ∈ bits.butlast, false ∈ bits, boolvar0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bitsand(bits.butlast, false ' + bits3.butlast) = false ' + bits3.butlast, # bits3 < # bits, ∀ bits, bits1. # bits1 < #(false + bits3) → # bits1 ≤ # bits → bits1 = [] ∨ bitsand(bits, bits1) = bits1 ∨ false ∈ bits ∨ bits1.first ⊦ bits3.butlast + (bits.last ∧ bits3.last)' = bits3, bits3 = [], bits = [], # bits = 1, false ∈ bits.butlast, false ∈ bits

Rule argument

Additional information

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