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Proof step 7 from proof referenced-adrs-nodups (previous step)

dups((adr ∈ ri ⊃ adr ';[]) + referenced-adrs(ax0, ri)) ⊦ dups(referenced-adrs(ax0, ri)), dups(ax0), adr ∈ ax0

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 1

Additional information

Used simplifier rules: referenced-adrs-in: ⊦ adr ∈ referenced-adrs(ax, ri) ↔ adr ∈ ax ∧ adr ∈ ri;% ⊦ [] + ax = ax; (s)%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[address-list-set]% ⊦ dups(adr ' + ax) ↔ dups(ax) ∨ adr ∈ ax; (ws)