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Proof step 3 from proof store-diff-union-elems-notin (previous step)

⊦ (gstore \ (ns ∪ elems nx))[n] = gstore[n], n ∈ ns, n ∈ nx

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ¬ n ∈ ns → (gstore \ ns)[n] = gstore[n]

Rule argument

spec: store+set, inst: group-store-ablock-store-nat-store name: Diff-get seq: ⊦ ¬ n ∈ ns → (gstore \ ns)[n] = gstore[n] subst: [n, ns, gstore] → [n, ns ∪ elems nx, gstore] flags: <all paths>

Additional information

Used simplifier rules: elems-def: ⊦ n ∈ elems nx ↔ n ∈ nx;% ⊦ n ∈ ns → (gstore \ ns)[n] = (∅)[n]; (s)% ⊦ ¬ n ∈ ns → (n ∈ ns ∪ ns0 ↔ n ∈ ns0); (s)%una: ⊦ (ns ∪ ns0) ∪ ns1 = ns ∪ ns0 ∪ ns1; from specification set-union[nat-set]%unc: ⊦ ns ∪ ns0 = ns0 ∪ ns; from specification set-union[nat-set] Used terms: n, ns ∪ elems nx, gstore