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Proof step 3 from proof lls-fold-head (previous step)

H[r0] = a0 × r3 ⊦ ((r0 => a × r1 * lls(r1, r2, x) * P))(H) ↔ (∃ r. (((r0 => a × r * lls(r, r2, x)) * P))(H)) ∧ True(r3 = r1)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
H[r0] = a0 × r3, ((r0 => a × r1 * lls(r1, r2, x) * P))(H) ⊦ (∃ r. (((r0 => a × r * lls(r, r2, x)) * P))(H)) ∧ True(r3 = r1)
Proof step 9:
H[r0] = a0 × r3, True(r3 = r1), ∃ r. (((r0 => a × r * lls(r, r2, x)) * P))(H) ⊦ ((r0 => a × r1 * lls(r1, r2, x) * P))(H)

Rule argument

pos: right 1

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