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Proof step 12 from proof bitsand-zero-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# bits = 1, false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < 1 → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], bits0 = true ', bits0 = false ', # bits0 = 1, false ∈ bitsand(bits, bits0)

Rule argument

Additional information

Used simplifier rules: ⊦ # bits = 1 ∧ bits ≠ true ' → bits = false '; (forward)% ⊦ # bits = 1 ∧ bits ≠ false ' → bits = true '; (forward)