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Proof step 3 from proof prefix-fwd-in (previous step)

n ∈ st0, dom st0 ⊆ dom st1, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist ⊦ n ∈ st1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0)
Proof step 5:
n ∈ st0, ∀ n. n ∈ st0 ∧ n ∈ st1 → st0[n].list ⊑ st1[n].list ∧ st0[n].ladrlist ⊑ st1[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st1[n].list = # st1[n].ladrlist, ∀ n. n ∈ dom st0 → n ∈ dom st1 ⊦ n ∈ st1

Rule argument

spec: set-basic, inst: nat-set name: Subset seq: ⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0) subst: [ns0, ns] → [dom st1, dom st0] flags: <all paths>

Additional information

Used terms: dom st1, dom st0