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Proof step 6 from proof unsubdel (
previous step
)
a0 ∈ s1, a1 ∈ s1, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ a0 ∈ s0, a0 ∈ s2, a ∈ s, a ∈ s1, a0 ∈ s
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: a0
Additional information
Used terms: a0