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Proof step 5 from proof nodes-remove-ginval-notin (previous step)

⊦ n ∈ gstore, ∀ adr. (adr ∈ nodes(gstore, ginval -- n) ↔ adr ∈ nodes(gstore, ginval)) ∧ (adr ∈ nodes(gstore, ginval -- n) → nodes(gstore, ginval -- n)[adr] = nodes(gstore, ginval)[adr])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ n ∈ gstore, (adr ∈ gstore ∧ adr.lnum = n → ¬ adr.lnum ∈ ginval) ∧ (adr ∈ gstore ∧ (adr.lnum = n ∨ ¬ adr.lnum ∈ ginval) → nodes(gstore, ginval -- n)[adr] = nodes(gstore, ginval)[adr])

Rule argument

Additional information

Used simplifier rules: nodes-gstore-in: ⊦ adr ∈ nodes(gstore, ginval) ↔ ¬ ¬(adr ∈ gstore ∧ ¬ adr.lnum ∈ ginval);% ⊦ n ∈ ns -- n0 ↔ n ≠ n0 ∧ n ∈ ns; (s)