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Proof step 12 from proof firstoc (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: [], y + a0 ' + z abort: F discard: F computed: F

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%app-08: ⊦ a ' + x = a ' + y ↔ x = y;%in-empty: ⊦ ¬ a ∈ []; Used terms: [], y + a0 ' + z