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Proof step 3 from proof prefixpoint-aux2 (
previous step
)
∀ S. f(S) ⊆ S → S0 ⊆ S ⊦ ∀ x. S0(x) → μ(f)(x)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 4
:
S0(x), ∀ S. f(S) ⊆ S → S0 ⊆ S ⊦ μ(f)(x)
Rule argument
Additional information