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Proof step 6 from proof cont-monotone (
previous step
)
∀ Sn. chain(Sn) → (∀ x. f(λ x0. ∃ n. Sn(n)(x0))(x) ↔ (∃ n. f(Sn(n))(x))) ⊦ ∀ S, S0. S ⊆ S0 → f(S) ⊆ f(S0)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 7
:
S0 ⊆ S, ∀ Sn. chain(Sn) → (∀ x. f(λ x0. ∃ n. Sn(n)(x0))(x) ↔ (∃ n. f(Sn(n))(x))) ⊦ f(S0) ⊆ f(S)
Rule argument
Additional information