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Proof step 13 from proof owner-ls1 (previous step)

rs1 ∩ rs0 = ∅, R[rs0, cf, af], ls(r, x)(mk(rs1, cf)) ⊦ af(r) = o, r = null, r ∈ rs1

User applied rule constructor cut

And obtained the following new goals

Proof step 14:
x = a + x0, rs1 ∩ rs0 = ∅, R[rs0, cf, af], ls(r, x)(mk(rs1, cf)) ⊦ af(r) = o, r = null, r ∈ rs1

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(H); (s)