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Proof step 5 from proof ls-unfold1 (previous step)

⊦ (∃ r0. (((r => a0 × r0 * ls(r0, x0)) * P))(H)) ↔ (∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ a0 ' + x0 = a ' + y), r = null

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ r0. (((r => a0 × r0 * ls(r0, x0)) * P))(H) ⊦ r = null, ∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ a0 ' + x0 = a ' + y
Proof step 8:
∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ a0 ' + x0 = a ' + y ⊦ r = null, ∃ r0. (((r => a0 × r0 * ls(r0, x0)) * P))(H)

Rule argument

pos: right 1

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