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Proof step 23 from proof dups-eqsize-02 (previous step)

# x0 = # y0 + # z, dups(y0 + a0 ' + z), ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0), a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, ∀ a1. a1 ∈ x0 → a1 ∈ y0 ∨ a1 ∈ z, a ∈ y0 ∨ a ∈ z → a ∈ x0, ∀ a2. a2 ∈ y0 ∨ a2 ∈ z → a2 ∈ x0 ⊦ a0 ∈ x0, dups(x0), dups(y0 + z), ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z

User applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: a0 abort: F discard: F computed: T

Additional information

Used simplifier rules: f-01: ⊦ dups(x + a ' + y) ↔ a ∈ x ∨ a ∈ y ∨ dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: a0