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Proof step 3 from proof ginval-down (previous step)

⊦ ∀ n. (n ∈ gstore ↓ \ ginval ↔ n ∈ (gstore \ ginval)↓) ∧ (n ∈ gstore ↓ \ ginval → (gstore ↓ \ ginval)[n] = (gstore \ ginval)↓[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (n ∈ gstore ↓ \ ginval ↔ n ∈ gstore \ ginval) ∧ (n ∈ gstore ↓ \ ginval → (gstore ↓ \ ginval)[n] = gstore ↓[n])

Rule argument

Additional information

Used simplifier rules: down-store-in: ⊦ n ∈ gstore ↓ ↔ n ∈ gstore;%down-store-at-diff: ¬ n ∈ ginval ⊦ (gstore \ ginval)↓[n] = gstore ↓[n];% ⊦ n ∈ gstore \ ns ↔ n ∈ gstore ∧ ¬ n ∈ ns; (ws)