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

bits0 = bits4 + boolvar0 ', bitsor(true ' + bits3, bits4) + (boolvar ∨ boolvar0)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits0 = [], # bits0 = 1, (boolvar ∨ boolvar0) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits3, bits4) ∧ (boolvar ∨ boolvar0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
bitsor(true ' + bits3, bits4) + (boolvar ∨ boolvar0)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits4 = [], (boolvar ∨ boolvar0) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits3, bits4) ∧ (boolvar ∨ boolvar0)

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)% ⊦ 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;% ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];