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Proof step 12 from proof del-lists-1 (previous step)

⊦ (∃ x. n = # x ∧ ls(r, x)(H)) ↔ (∃ x. ls(r, x)(H) ∧ n = # x), r = null

User applied rule case distinction

And obtained the following new goals

Proof step 13:
∃ x. n = # x ∧ ls(r, x)(H) ⊦ r = null, ∃ x. ls(r, x)(H) ∧ n = # x
Proof step 15:
∃ x. ls(r, x)(H) ∧ n = # x ⊦ r = null, ∃ x. n = # x ∧ ls(r, x)(H)

Rule argument

pos: right 1

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