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Proof step 2 from proof in-get (previous step)

a ∈ x ⊦ ∃ n. n < # x ∧ x[n] = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 4:
∃ y, z. x = y + (a + z) ⊦ ∃ n. n < # x ∧ x[n] = a

Rule argument

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

Additional information

Used terms: x, a