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Proof step 5 from proof ll-trav1-next (previous step)

H[r1] = a × r, ((lls(r0, r1, x0) * lls(r1, null, x1) * P))(H) ⊦ r1 = null, x1 = x1.head ' + x1.tail ∧ ((lls(r0, r, x0 + x1.head ') * lls(r, null, x1.tail) * P))(H)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: list-data name: elim-list-c seq: x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y subst: [x, a, y] → [x1, a0, y]

Additional information

Used simplifier rules: lls-base: ⊦ lls(r0, r1, []) = (emp ∧ ⌜r0 = r1⌝ );% ⊦ ⌜false⌝ = false; (s)% ⊦ (false ∧ P) = false; (s)%and-ass: ⊦ ((P0 ∧ P1) ∧ P2) = (P0 ∧ P1 ∧ P2); from specification heap-sep%and-com: ⊦ (P0 ∧ P1) = (P1 ∧ P0); from specification heap-sep% ⊦ false * P = false; (s)%sep-ass: ⊦ (P0 * P1) * P2 = P0 * P1 * P2; from specification heap-sep%sep-com: ⊦ P0 * P1 = P1 * P0; from specification heap-sep% ⊦ ¬ false(H); (s)%x ≠ [] ⊦ a = x.head ∧ y = x.tail ↔ x = a + y; (elim) Used terms: x1, a0, y