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Proof step 5 from proof lls-unfold-head (previous step)

⊦ (∃ r. (((r0 => a0 × r * lls(r, r2, x0)) * P))(H)) ↔ (∃ a, y, r. ((r0 => a × r * lls(r, r2, y) * P))(H) ∧ a0 ' + x0 = a ' + y)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∃ r. (((r0 => a0 × r * lls(r, r2, x0)) * P))(H) ⊦ ∃ a, y, r. ((r0 => a × r * lls(r, r2, y) * P))(H) ∧ a0 ' + x0 = a ' + y
Proof step 8:
∃ a, y, r. ((r0 => a × r * lls(r, r2, y) * P))(H) ∧ a0 ' + x0 = a ' + y ⊦ ∃ r. (((r0 => a0 × r * lls(r, r2, x0)) * P))(H)

Rule argument

pos: right 1

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