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Proof step 4 from proof remove-refsize0-down (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ n ∈ remove-refsize0(gstore) ↔ n ∈ gstore ∧ gstore[n].ref-size ≠ 0
Proof step 6:
⊦ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 → remove-refsize0(gstore)[n] ↓ = remove-refsize0(gstore ↓)[n])

Rule argument

spec: ablock-store-inv, inst: group-store name: in-remove-refsize0 seq: ⊦ n ∈ remove-refsize0(gstore) ↔ n ∈ gstore ∧ gstore[n].ref-size ≠ 0 subst: [gstore, n] → [gstore, n] flags: <all paths>

Additional information

Used terms: gstore, n