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Proof step 5 from proof fixpoint (
previous step
)
monotone(f), monotone(f) ⊦ true ∧ μ(f) ⊆ f(μ(f))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 6
:
monotone(f) ⊦ μ(f) ⊆ f(μ(f))
Rule argument
Additional information