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Proof step 23 from proof bitsor-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
bitsor(true ' + bits3, bits4) + (boolvar ∨ boolvar0)' = false ' + bits1, (boolvar ∨ boolvar0) ∈ bits1, false ∈ bitsor(true ' + bits3, bits4), ∀ bits1, bits, bits2. # bits2 < #(bits4 + boolvar0 ') → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits1 = [], bits4 = [], (boolvar ∨ boolvar0) ∧ ¬ true ∈ bits1

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;%⊦ boolvar ' + bits = bits0 + boolvar0 ' ∧ bits ≠ [] → boolvar ∈ bits0 ∧ boolvar0 ∈ bits; (forward)