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Proof step 6 from proof in-def (previous step)

∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z) ⊦ ∃ y, z. a0 ' + x0 = y + a0 ' + z

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], x0 abort: F discard: T computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%app-01: ⊦ x + y = x + z ↔ y = z; Used terms: [], x0