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Proof step 6 from proof dups-lesize (previous step)

# y ≤ # x0, a ∈ y, ∀ a0. a0 ∈ x0 → a0 ∈ y, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0) ⊦ a ∈ x0, dups(x0), dups(y)

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:
# y ≤ # x0, ∃ y0, z. y = y0 + (a + z), ∀ a0. a0 ∈ x0 → a0 ∈ y, ∀ y0. # y0 < # x0 ∧ (∀ a. a ∈ x0 → a ∈ y0) → x0 = [] ∨ dups(y0) ⊦ a ∈ x0, dups(x0), dups(y)

Rule argument

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

Additional information

Used terms: y, a