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

((r0 => a × r1 * lls(r1, r2, x) * P))(H) ⊦ (∃ r. (((r0 => a × r * lls(r, r2, x)) * P))(H)) ∧ True(true)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
((r0 => a × r1 * lls(r1, r2, x) * P))(H) ⊦ ∃ r. (((r0 => a × r * lls(r, r2, x)) * P))(H)
Proof step 7:
((r0 => a × r1 * lls(r1, r2, x) * P))(H) ⊦ True(true)

Rule argument

pos: right 1

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