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Proof step 3 from proof bitsand-zero-in (previous step)

false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits0 = [] ∨ bits = [] ∨ false ∈ bitsand(bits, bits0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
false ∈ bits ∨ false ∈ bits0, ∀ bits0, bits1. # bits1 < # bits → false ∈ bits1 ∨ false ∈ bits0 → bits0 = [] ∨ bits1 = [] ∨ false ∈ bitsand(bits1, bits0) ⊦ bits = [], bits0 = [], false ∈ bitsand(bits, bits0)

Proof step 3 from proof bitsand-zero-in (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information