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Proof step 18 from proof down-except-refsize (previous step)

gstore1 ↓[n] = groupblock(gndx, lax, m), (∅)[n] = groupblock(gndx0, lax0, m0), n ∈ gstore1 ↓, ∀ n. (n ∈ gstore1 ↔ n ∈ gstore0) ∧ gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist ⊦ n ∈ gstore0, ¬ n ∈ gstore1 ∧ gndx = gndx0 ∧ lax = lax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
(∅)[n] = groupblock(gndx0, lax0, m0), gstore1[n] ↓ = groupblock(gndx, lax, m), n ∈ gstore1, ∀ n. (n ∈ gstore1 ↔ n ∈ gstore0) ∧ gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist ⊦ n ∈ gstore0

Rule argument

Additional information

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