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Proof step 18 from proof bitsor-first-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bitsor(false ' + bits0.butlast, false ' + bits2.butlast) + (bits0.last ∨ bits2.last)' = true ' + bits1, ∀ bits1, bits2, bits. # bits < # bits0 → bitsor(false ' + bits, false ' + bits2) = true ' + bits1 → false ⊦ bits2 = [], bits0 = [], ¬ true ∈ bitsor(false ' + bits0.butlast, false ' + bits2.butlast) ∧ ¬ bits0.last ∧ ¬ bits2.last, ¬ bits0.last ∧ ¬ bits2.last ∧ ¬ false ∈ bits1

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;% ⊦ # bits = 0 ↔ bits = [];% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + 1 ≠ 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ # boolvar ' = 0 +1;% ⊦ #(bits + bits0) = # bits + # bits0;% ⊦ bits + boolvar = bits + boolvar ';