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Proof step 4 from proof reachnext (previous step)

⊦ (∃ x. path(r0 ' + x, H) ∧ (r0 ' + x).last = H[r0].nxt) ↔ r0 ≠ null ∧ r0 ∈ H ∧ H[r0].nxt ∈ H ∧ H[r0].nxt ≠ null

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∃ x. path(r0 ' + x, H) ∧ (r0 ' + x).last = H[r0].nxt ⊦ r0 ≠ null ∧ r0 ∈ H ∧ H[r0].nxt ∈ H ∧ H[r0].nxt ≠ null
Proof step 7:
r0 ≠ null ∧ r0 ∈ H ∧ H[r0].nxt ∈ H ∧ H[r0].nxt ≠ null ⊦ ∃ x. path(r0 ' + x, H) ∧ (r0 ' + x).last = H[r0].nxt

Rule argument

pos: right 1

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