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Proof step 11 from proof in-diff (previous step)

(st \ ns)[m] = mkblock(x, lax, m2), st[m] = mkblock(x0, lax0, m3) ⊦ m ∈ st \ ns ∧ @(m0, m1) ∈ lax ↔ m ∈ st ∧ @(m0, m1) ∈ lax0 ∧ ¬ m ∈ ns

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ n ∈ st \ ns ↔ n ∈ st ∧ ¬ n ∈ ns

Rule argument

spec: store+set, inst: ablock-store-nat-store name: Diff-ext seq: ⊦ n ∈ st \ ns ↔ n ∈ st ∧ ¬ n ∈ ns subst: [ns, st, n] → [ns, st, m] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ¬ n ∈ ns → (st \ ns)[n] = st[n]; (s)%⊦ mkblock(x, lax, m) = mkblock(x0, lax0, m0) ↔ x = x0 ∧ lax = lax0 ∧ m = m0; (ws) Used terms: ns, st, m