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Proof step 4 from proof inv-diff-n (previous step)

st[n] = mkblock(x, lax, m), n ∈ st, ∀ n0. n0 ∈ st \ ns → # (st \ ns)[n0].list = # (st \ ns)[n0].ladrlist ∧ inv-flashsize((st \ ns)[n0].list, (st \ ns)[n0].ladrlist) ∧ ¬ dups((st \ ns)[n0].ladrlist) ⊦ n ∈ ns, inv-flashsize(x, lax) ∧ ¬ dups(lax)

User applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ n ∈ st \ ns ↔ n ∈ st ∧ ¬ n ∈ ns; (ws)% ⊦ ¬ n ∈ ns → (st \ ns)[n] = st[n]; (s)% ⊦ mkblock(x, lax, m).list = x; (s)% ⊦ mkblock(x, lax, m).ladrlist = lax; (s) Used terms: n