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Proof step 15 from proof prefix-append-notin (previous step)

st0[n0] = mkblock(x, lax, m), st[n0] = mkblock(x0, lax0, m0), n0 ∈ st0, n0 ∈ st, 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

Heuristic quantifier applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: n0

Additional information

Used simplifier rules: ⊦ mkblock(x, lax, m).list = x; (s)% ⊦ mkblock(x, lax, m).ladrlist = lax; (s) Used terms: n0