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Proof step 8 from proof in-def (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 9:
a ∈ x0, a ∈ x0 → (∃ y, z. x0 = y + a ' + z) ⊦ a = a0, ∃ y, z. a0 ' + x0 = y + a ' + z

Rule argument

pos: left 2 terms: a abort: F discard: T computed: T

Additional information

Used terms: a