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Proof step 4 from proof self-reachable (previous step)

⊦ (∃ x. path(r ' + x, H) ∧ (r ' + x).last = r) ↔ r ≠ null ∧ r ∈ H

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ x. path(r ' + x, H) ∧ (r ' + x).last = r ⊦ r ≠ null ∧ r ∈ H
Proof step 7:
r ≠ null ∧ r ∈ H ⊦ ∃ x. path(r ' + x, H) ∧ (r ' + x).last = r

Rule argument

pos: right 1

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