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Proof step 6 from proof dups-firstoc-rew (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 7:
a0 ∈ y ⊦ dups(y), a ' + y + a0 ' + z = (a ' + y) + a0 ' + z ∧ a0 ∈ a ' + y ∧ ¬ dups(a ' + y), ∃ y0, a1, z0. a ' + y + a0 ' + z = y0 + a1 ' + z0 ∧ a1 ∈ y0 ∧ ¬ dups(y0)

Rule argument

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

Additional information

Used terms: a ' + y, a0, z