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Proof step 16 from proof bitsor-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
bitsor(true ' + bits3, bits0.butlast) + (boolvar ∨ bits0.last)' = false ' + bits1, ∀ bits1, bits, bits2. # bits2 < # bits0 → bitsor(true ' + bits, bits2) = false ' + bits1 → bits2 = [] ⊦ bits0 = [], # bits0 = 1, (boolvar ∨ bits0.last) ∧ ¬ true ∈ bits1, ¬ false ∈ bitsor(true ' + bits3, bits0.butlast) ∧ (boolvar ∨ bits0.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 + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];