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Proof step 9 from proof prefixpoint-aux2 (
previous step
)
f(S) ⊆ S, S0(x), S0 ⊆ S ⊦ S(x)
User applied rule insert rewrite lemma
And obtained the following new goals
Proof step 10
:
⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x))
Proof step 11
:
f(S) ⊆ S, S0(x), ∀ x. S0(x) → S(x) ⊦ S(x)
Rule argument
spec: inst: name: subseteq-def seq: ⊦ S ⊆ S0 ↔ (∀ x. S(x) → S0(x)) subst: [S0, S] → [S, S0] flags: rewrite, current
Additional information
Used terms: S, S0