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Proof step 11 from proof dups-eqsize-01 (previous step)

# x0 = # y0 + # z, a1 ∈ x0, ∀ y0, a1. # x0 = # y0 ∧ a1 ∈ y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → a1 ∈ x0 ∨ dups(y0), a ∈ y0 ∨ a ∈ z, ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z ⊦ a = a0, a0 ∈ x0, dups(x0), a ∈ x0, a0 ∈ y0, a0 ∈ z, dups(y0 + z), a1 ∈ y0, a1 ∈ z

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 5 terms: a1

Additional information

Used terms: a1