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Proof step 11 from proof f-01 (
previous step
)
a ∈ y, ∀ a, y. a ∈ x0 ∨ a ∈ y ∨ dups(x0 + y) → dups(x0 + a ' + y) ∨ ¬ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0, a0 ∈ x0, a0 ∈ y, dups(x0 + a ' + y)
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: a, y
Additional information
Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a, y