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Proof step 8 from proof notin-add-diff-dom-nodes (previous step)

adr0 = adr ∨ adr0 ∈ as \ dom nodes(st) ⊦ adr ∈ st, (adr0 = adr ∨ adr0 ∈ as) ∧ ¬ adr0 ∈ st

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: in-nodes: ⊦ adr ∈ nodes(st) ↔ adr ∈ st;% ⊦ adr ∈ as \ as0 ↔ adr ∈ as ∧ ¬ adr ∈ as0; (ws)% ⊦ adr ∈ dom ast ↔ adr ∈ ast; (s)