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Proof step 10 from proof firstoc (previous step)

∀ a1. a1 ∈ y ∨ a1 = a ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ a ∈ y, ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0

User applied rule exists right

And obtained the following new goals

Proof step 11:
∀ a1. a1 ∈ y ∨ a1 = a ∨ a1 ∈ z → (∃ y0, z0. y + a ' + z = y0 + a1 ' + z0 ∧ ¬ a1 ∈ y0) ⊦ a ∈ y, a0 ' + y + a ' + z = (a0 ' + y) + a ' + z ∧ ¬ a ∈ a0 ' + y, ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0 ∧ ¬ a ∈ y0

Rule argument

pos: right 2 terms: a0 ' + y, z abort: F discard: F computed: T

Additional information

Used terms: a0 ' + y, z