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Proof step 9 from proof ginval-down (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
n ∈ ginval ⊦ # ginval = 0, ginval = ∅, n ∈ gstore, (n ∈ gstore ↓ \ ginval ↔ n ∈ gstore \ ginval) ∧ (n ∈ gstore ↓ \ ginval → (∅)[n] = gstore ↓[n])

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ns ≠ ∅ → # ns ≠ 0; (forward) from specification set-basic[nat-set]%ne-01: ⊦ n ∈ ns → ns ≠ ∅; (forward) from specification set-basic[nat-set]% ⊦ n ∈ ns → (gstore \ ns)[n] = (∅)[n]; (s)