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Proof step 3 from proof at-explicit-diff (previous step)

# x0 = # lax0, st[n] = mkblock(x0 + a ' + x1, lax0 + @(m0, m1) ' + lax1, m2), 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) ⊦ st[@(n, m0, m1)] = a, n ∈ ns

User applied rule all left

And obtained the following new goals

Proof step 4:
# x0 = # lax0, st[n] = mkblock(x0 + a ' + x1, lax0 + @(m0, m1) ' + lax1, m2), n ∈ st, n ∈ st \ ns → # (st \ ns)[n].list = # (st \ ns)[n].ladrlist ∧ inv-flashsize((st \ ns)[n].list, (st \ ns)[n].ladrlist) ∧ ¬ dups((st \ ns)[n].ladrlist) ⊦ st[@(n, m0, m1)] = a, n ∈ ns

Rule argument

pos: left 4 terms: n abort: F discard: T computed: F

Additional information

Used terms: n