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Proof step 6 from proof prefix-in (previous step)

st0[m] = mkblock(x, lax, m2), @(m0, m1) ∈ lax, m ∈ 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 ⊦ @(m, m0, m1) ∈ st1

User applied rule all left

And obtained the following new goals

Proof step 7:
st0[m] = mkblock(x, lax, m2), m ∈ st0, @(m0, m1) ∈ lax, dom st0 ⊆ dom st1, m ∈ st0 ∧ m ∈ st1 → st0[m].list ⊑ st1[m].list ∧ st0[m].ladrlist ⊑ st1[m].ladrlist ∧ # st0[m].list = # st0[m].ladrlist ∧ # st1[m].list = # st1[m].ladrlist, ∀ 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 ⊦ @(m, m0, m1) ∈ st1

Rule argument

pos: left 5 terms: m abort: F discard: F computed: T

Additional information

Used terms: m