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Proof step 9 from proof bitsand-onein (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ bits0, bits, bits2. # bits2 < # bits1 → true → bits2 = [] ∨ (true ∈ bitsand(bits + false ' + bits0, bits2) ↔ true ∈ bitsand(false ' + bits0, bits2)) ∨ true ∈ bits ⊦ true ∈ bitsand(bits + false ' + bits0, bits1) ↔ true ∈ bitsand(false ' + bits0, bits1), bits1 = [], bits1 = true ', true ∈ bits, bits0 ≠ [] ∧ # bits1 ≠ 1

Rule argument

Additional information

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