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Proof step 3 from proof remove-refsize0-down (previous step)

⊦ ∀ n. (n ∈ remove-refsize0(gstore) ↓ ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ remove-refsize0(gstore) ↓ → remove-refsize0(gstore) ↓[n] = remove-refsize0(gstore ↓)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (n ∈ remove-refsize0(gstore) ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ remove-refsize0(gstore) → remove-refsize0(gstore)[n] ↓ = remove-refsize0(gstore ↓)[n])

Rule argument

Additional information

Used simplifier rules: down-store-in: ⊦ n ∈ gstore ↓ ↔ n ∈ gstore;%down-store-at: ⊦ n ∈ gstore → gstore ↓[n] = gstore[n] ↓;