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Proof step 5 from proof referenced-adrs-in (previous step)

adr ∈ (adr0 ∈ ri ⊃ adr0 ';[]), ∀ adr, ri. adr ∈ referenced-adrs(ax0, ri) ↔ adr ∈ ax0 ∧ adr ∈ ri ⊦ (adr = adr0 ∨ adr ∈ ax0) ∧ adr ∈ ri

Heuristic cut applied rule cut formula

And closed the goal!

Rule argument

fma: adr0 ∈ ri

Additional information

Used simplifier rules: ⊦ ¬ adr ∈ []; (s)% ⊦ adr ∈ adr0 ' ↔ adr = adr0; (s) Used formulas: adr0 ∈ ri