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Proof step 5 from proof intempty (
previous step
)
a ∈ y, a ∈ x, ∀ a0. ¬ a0 ∈ list2set(x) ∩ list2set(y) ⊦
User applied rule all left
And obtained the following new goals
Proof step 6
:
a ∈ y, a ∈ x, ¬ a ∈ list2set(x) ∩ list2set(y) ⊦
Rule argument
pos: left 3 terms: a abort: F discard: T computed: T
Additional information
Used terms: a