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Proof step 4 from proof lls-unfold-head (previous step)

x = a0 + x0 ⊦ ((lls(r0, r2, x) * P))(H) ↔ (∃ a, y, r1. ((r0 => a × r1 * lls(r1, r2, y) * P))(H) ∧ x = a ' + y), x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (∃ 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)

Rule argument

Additional information

Used simplifier rules: lls-rec: ⊦ lls(r, r1, a ' + x) = _∃(λ r0. r => a × r0 * lls(r0, r1, x));%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep% ⊦ ((_∃ rP * P))(H) ↔ (∃ r. ((rP(r) * P))(H)); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' + x ≠ []; (s)% ⊦ a + x = a ' + x; (s)