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Proof step 31 from proof nu-is-UN (
previous step
)
S(x0), S ⊆ f(S), S(x) ⊦ f(S)(x0)
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 32
:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 33
:
S(x0), ∀ x. S(x) → f(S)(x), S(x) ⊦ f(S)(x0)
Rule argument
spec: inst: name: subseteq-def seq: ⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x)) subst: [S0, S] → [f(S), S] flags: rewrite, current
Additional information
Used terms: f(S), S