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Proof step 9 from proof dups-eqsize-01 (previous step)

# x0 = # y0 + # z, ∀ 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)

User applied rule all left

And obtained the following new goals

Proof step 10:
# x0 = # y0 + # z, # x0 = #(y0 + z) ∧ a ∈ y0 + z ∧ (∀ a. a ∈ x0 → a ∈ y0 + z) → a ∈ x0 ∨ dups(y0 + z), ∀ 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)

Rule argument

pos: left 2 terms: y0 + z, a abort: F discard: F computed: T

Additional information

Used terms: y0 + z, a