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Proof step 12 from proof prefix-rem (previous step)

st[n0] = mkblock(x0, lax0, m0), st0[n0] = mkblock(x, lax, m), n0 ∈ st, n0 ∈ st0, ∀ n. n ∈ st0 \ ns ∧ n ∈ st \ ns → (st0 \ ns)[n].list ⊑ (st \ ns)[n].list ∧ (st0 \ ns)[n].ladrlist ⊑ (st \ ns)[n].ladrlist ∧ # (st0 \ ns)[n].list = # (st0 \ ns)[n].ladrlist ∧ # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ⊦ n0 = n, n0 ∈ ns, x ⊑ x0 ∧ lax ⊑ lax0 ∧ # x = # lax ∧ # x0 = # lax0

User applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: n0 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: n0