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Proof step 11 from proof dups-firstoc-rew (previous step)

a0 ∈ y0 ∨ a0 = a ∨ a0 ∈ z0 ⊦ a ∈ y0, a ∈ z0, dups(y0 + z0), ∃ y1, a1, z1. a ' + (y0 + a ' + z0) + a0 ' + z = y1 + a1 ' + z1 ∧ a1 ∈ y1 ∧ ¬ dups(y1)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a ' + y0, a, z0 + a0 ' + z abort: F discard: T computed: F

Additional information

Used simplifier rules: rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x));%app-05: ⊦ dups(x) → dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list%in-10: ⊦ ¬ a ∈ x → (a ∈ b ' + x ↔ a = b); from specification list Used terms: a ' + y0, a, z0 + a0 ' + z