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Proof step 39 from proof bitsor-sgn-in (
previous step
)
bits0 ≠ [] ∧ false ' = false ' ⊦ bits = [], bits0 = [], boolvar ∈ bits0 ↔ boolvar ∈ bitsor(bits, bits0), true ∈ bits
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 40
:
⊦ bits = [], boolvar ∈ bits0 ↔ boolvar ∈ bitsor(bits, bits0), bits0 = [], true ∈ bits
Rule argument
Additional information