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Proof step 10 from proof unique-put (
previous step
)
∀ a, d. unique_al(ax0[a, d]) ↔ unique_al(ax0) ⊦ a = a0, unique_al(ax0[a, d]) ↔ unique_al(ax0), a0 ∈ ax0
User applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: a, d abort: F discard: T computed: T
Additional information
Used terms: a, d