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Proof step 5 from proof new-not-new (
previous step
)
new(ax0, fs), adr ∈ fs, adr ∈ ax0, ∀ fs0, adr1. ¬ new(ax0, fs0) ∨ ¬ adr1 ∈ ax0 ∨ ¬ adr1 ∈ fs0 ⊦ adr0 ∈ fs
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 4 terms: fs, adr
Additional information
Used terms: fs, adr