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Proof step 4 from proof reachable-nodups-def (previous step)

⊦ (∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0) ↔ (∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0 ∧ ¬ r ∈ x ∧ ¬ dups(x))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0 ⊦ ∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0 ∧ ¬ r ∈ x ∧ ¬ dups(x)
Proof step 22:
∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0 ∧ ¬ r ∈ x ∧ ¬ dups(x) ⊦ ∃ x. path(r ' + x, H) ∧ (r ' + x).last = r0

Rule argument

pos: right 1

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