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Proof step 18 from proof gstore-dom-eq-hole (previous step)

n0 ∈ gstore0 \ (∅ ++ n) ↔ n0 ∈ gstore1 \ (∅ ++ n), (gstore0 \ (∅ ++ n))[n0] = (gstore1 \ (∅ ++ n))[n0], n0 ∈ gstore0, n0 ∈ gstore1, n ∈ gstore0 ⊦ gstore1[n0] = gstore0[n, gstore1[n]][n0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ n ≠ n0 → gstore[n0, gblock][n] = gstore[n]
Proof step 20:
n0 ∈ gstore0 \ (∅ ++ n) ↔ n0 ∈ gstore1 \ (∅ ++ n), (gstore0 \ (∅ ++ n))[n0] = (gstore1 \ (∅ ++ n))[n0], n ∈ gstore0, n0 ∈ gstore1, n0 ∈ gstore0 ⊦ gstore1[n0] = gstore0[n0], n0 = n

Rule argument

spec: store, inst: group-store-ablock-store-nat-store name: At-other seq: ⊦ n ≠ n0 → gstore[n0, gblock][n] = gstore[n] subst: [n, gblock, n0, gstore] → [n0, gstore1[n], n, gstore0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ n ∈ gstore \ ns ↔ n ∈ gstore ∧ ¬ n ∈ ns; (ws)% ⊦ n ∈ ∅ ++ n0 ↔ n = n0; (s)% ⊦ gstore[n, gblock][n] = gblock; (s) Used terms: n0, gstore1[n], n, gstore0