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Proof step 5 from proof nodes-no-change-eq (previous step)

∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist) ⊦ ∀ adr. (adr ∈ nodes(gstore0, ginval) ↔ adr ∈ nodes(gstore1, ginval)) ∧ (adr ∈ nodes(gstore0, ginval) → nodes(gstore0, ginval)[adr] = nodes(gstore1, ginval)[adr])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist) ⊦ (¬ adr.lnum ∈ ginval → (adr ∈ gstore0 ↔ adr ∈ gstore1)) ∧ (adr ∈ gstore0 ∧ ¬ adr.lnum ∈ ginval → nodes(gstore0, ginval)[adr] = nodes(gstore1, ginval)[adr])

Rule argument

Additional information

Used simplifier rules: nodes-gstore-in: ⊦ adr ∈ nodes(gstore, ginval) ↔ ¬ ¬(adr ∈ gstore ∧ ¬ adr.lnum ∈ ginval);