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Proof step 4 from proof firstoc (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 5:
a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), ∀ a. a ∈ x0 → (∃ y, z. x0 = y + a ' + z ∧ ¬ a ∈ y), a = a0 ∨ a ∈ x0 ⊦ ∃ y, z. a0 ' + x0 = y + a ' + z ∧ ¬ a ∈ y

Rule argument

pos: left 1 terms: a abort: F discard: F computed: T

Additional information

Used terms: a