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Proof step 18 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 ⊦ a0 ∈ x0, dups(x0), dups(y0 + z), ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z

User applied rule insert lemma

And obtained the following new goals

Proof step 19:
# x = # y, ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ ∀ a. a ∈ y → a ∈ x
Proof step 20:
# 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 ⊦ # x0 = #(y0 + z) ∧ ¬ dups(x0) ∧ ¬ dups(y0 + z) ∧ (∀ a. a ∈ x0 → a ∈ y0 + z), a0 ∈ x0, dups(x0), dups(y0 + z), ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z
Proof step 22:
∀ a. a ∈ y0 + z → a ∈ x0, # 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 ⊦ a0 ∈ x0, dups(x0), dups(y0 + z), ¬ a0 ∈ y0 ∧ ¬ a0 ∈ z

Rule argument

spec: inst: name: dups-eqsize-01 seq: # x = # y, ¬ dups(x), ¬ dups(y), ∀ a. a ∈ x → a ∈ y ⊦ ∀ a. a ∈ y → a ∈ x subst: [x, y] → [x0, y0 + z] flags: current, []

Additional information

Used terms: x0, y0 + z