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Proof step 9 from proof compl-monotone (
previous step
)
S0 ⊆ S, ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0) ⊦ ∀ x. compl(f)(S0)(x) → compl(f)(S)(x)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 10
:
compl(f)(S0)(x), S0 ⊆ S, ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0) ⊦ compl(f)(S)(x)
Rule argument
Additional information