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Proof step 10 from proof prefix-append-notin (previous step)

st0[n] = mkblock(x, lax, m), n ∈ st0, dom st0 ⊆ dom st, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist ⊦ n ∈ st, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ ns ⊆ ns0 ↔ (∀ n. n ∈ ns → n ∈ ns0)
Proof step 12:
st0[n] = mkblock(x, lax, m), n ∈ st0, ∀ n. n ∈ st0 ∧ n ∈ st → st0[n].list ⊑ st[n].list ∧ st0[n].ladrlist ⊑ st[n].ladrlist ∧ # st0[n].list = # st0[n].ladrlist ∧ # st[n].list = # st[n].ladrlist, ∀ n. n ∈ dom st0 → n ∈ dom st ⊦ n ∈ st, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

Rule argument

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

Additional information

Used terms: dom st, dom st0