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Proof step 3 from proof ls-unfold1 (previous step)

⊦ ((ls(r, x) * P))(H) ↔ (∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ x = a ' + y), r = null

User applied rule constructor cut

And obtained the following new goals

Proof step 4:
x = a0 + x0 ⊦ ((ls(r, x) * P))(H) ↔ (∃ a, y, r0. ((r => a × r0 * ls(r0, y) * P))(H) ∧ x = a ' + y), r = null

Rule argument

var: x

Additional information

Used simplifier rules: ls-base: ⊦ ls(r, []) = (emp ∧ ⌜r = null⌝ );%list freely generated by [], +;;% ⊦ ⌜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)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list