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Proof step 4 from proof in-rec (previous step)

⊦ (∃ y, z. b ' + x = y + (a + z)) ↔ a = b ∨ a ∈ x

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ x ↔ (∃ y, z. x = y + (a + z))
Proof step 6:
⊦ (∃ y, z. b ' + x = y + (a + z)) ↔ a = b ∨ (∃ y, z. x = y + (a + z))

Rule argument

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

Additional information

Used terms: x, a