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Proof step 6 from proof dups-eqsize-02 (previous step)

# x0 + 1 = # y, a0 ∈ y, ∀ a1. a1 ∈ x0 → a1 ∈ y, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0) ⊦ a0 ∈ x0, dups(x0), ¬ dups(y) ∧ (a ∈ y → a = a0 ∨ a ∈ x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 8:
# x0 + 1 = # y, ∃ y0, z. y = y0 + (a0 + z), ∀ a1. a1 ∈ x0 → a1 ∈ y, ∀ y0, a1. # x0 = # y0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → ¬ dups(y0) ∧ (a1 ∈ y0 → a1 ∈ x0) ⊦ a0 ∈ x0, dups(x0), ¬ dups(y) ∧ (a ∈ y → a = a0 ∨ a ∈ x0)

Rule argument

spec: list inst: name: In seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)) subst: [x, a] → [y, a0] flags: rewrite

Additional information

Used terms: y, a0