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Proof step 2 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(H[r0].nxt = r1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
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)

Rule argument

spec: cell name: elim seq: ⊦ a = ce.val ∧ r = ce.nxt ↔ ce = a × r subst: [ce, a, r] → [H[r0], a0, r3]

Additional information

Used simplifier rules: ⊦ a = ce.val ∧ r = ce.nxt ↔ ce = a × r; (elim) Used terms: H[r0], a0, r3