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Proof step 7 from proof remove-refsize0-down (previous step)

⊦ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 → gstore[n] ↓ = remove-refsize0(gstore ↓)[n])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
gstore[n] = groupblock(gndx, lax, m) ⊦ (n ∈ gstore ∧ m ≠ 0 ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ gstore ∧ m ≠ 0 → gstore[n] ↓ = remove-refsize0(gstore ↓)[n])

Rule argument

spec: ablock, inst: group-store name: elim seq: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m) subst: [gblock, gndx, lax, m] → [gstore[n], gndx, lax, m]

Additional information

Used simplifier rules: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m); (elim) Used terms: gstore[n], gndx, lax, m