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Proof step 4 from proof compl-monotone (
previous step
)
monotone(f) ⊦ ∀ S, S0. S ⊆ S0 → compl(f)(S) ⊆ compl(f)(S0)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 5
:
S0 ⊆ S, monotone(f) ⊦ compl(f)(S0) ⊆ compl(f)(S)
Rule argument
Additional information