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Proof step 23 from proof bitsand-onein (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ bits1 = [], true ∈ bitsand(bits + false ' + bits3 + boolvar ', bits1) ↔ true ∈ bitsand(false ' + bits3, bits1.butlast) ∨ boolvar ∧ bits1.last, # bits1 = 1, true ∈ bits

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ bits + boolvar ' ≠ [];