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Proof step 8 from proof reach-n-path-n (previous step)

⊦ r ≠ null ∧ H[r].nxt ≠ null ∧ r ∈ H ∧ H[r].nxt ∈ H ∧ H[r].nxt = r0 ↔ (∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 2)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
r ≠ null ∧ H[r].nxt ≠ null ∧ r ∈ H ∧ H[r].nxt ∈ H ∧ H[r].nxt = r0 ⊦ ∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 2
Proof step 11:
∃ x. path(x, H) ∧ hd x = r ∧ x.last = r0 ∧ # x = 2 ⊦ r ≠ null ∧ H[r].nxt ≠ null ∧ r ∈ H ∧ H[r].nxt ∈ H ∧ H[r].nxt = r0

Rule argument

pos: right 1

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