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Proof step 6 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 → remove-refsize0(gstore)[n] ↓ = remove-refsize0(gstore ↓)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 ↔ n ∈ remove-refsize0(gstore ↓)) ∧ (n ∈ gstore ∧ gstore[n].ref-size ≠ 0 → gstore[n] ↓ = remove-refsize0(gstore ↓)[n])

Rule argument

Additional information

Used simplifier rules: ⊦ n ∈ gstore ∧ gstore[n].ref-size ≠ 0 → remove-refsize0(gstore)[n] = gstore[n]; (s)