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Proof step 5 from proof last-ex (previous step)

a ∈ y0 ⊦ a = a0, ∃ y, z. y0 + a0 ' = y + a ' + z ∧ z ≠ []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 7:
∃ y, z. y0 = y + (a + z) ⊦ a = a0, ∃ y, z. y0 + a0 ' = y + a ' + z ∧ z ≠ []

Rule argument

spec: list inst: name: In seq: ⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z)) subst: [x, a] → [y0, a] flags: rewrite

Additional information

Used terms: y0, a