Proof step 1 from proof ls-null (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ls(null, x) ≠ (emp ∧ ⌜x = []⌝ ) ⊦

Rule argument

Additional information

Proof step 2 from proof ls-null (previous step) (back to tree)

User applied rule constructor cut

And obtained the following new goals

• Proof step 3:

  x = a + x0, ls(null, x) ≠ (emp ∧ ⌜x = []⌝ ) ⊦

Rule argument

Additional information

Used simplifier rules: ls-base: ⊦ ls(r, []) = (emp ∧ ⌜r = null⌝ ); list freely generated by [], +;; ⊦ ⌜true⌝ = true; (s) ⊦ (true ∧ hP) = hP; (s) and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep

Proof step 3 from proof ls-null (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  _∃(λ r0. false) ≠ false ⊦

Rule argument

Additional information

Used simplifier rules: ls-rec: ⊦ ls(r, a ' + x) = _∃(λ r0. r => a × r0 * ls(r0, x)); and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep ⊦ (false ∧ hP) = false; (s) ⊦ ⌜false⌝ = false; (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a ' + x ≠ []; (s) ⊦ a + x = a ' + x; (s) sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep ⊦ false * hP = false; (s) ⊦ null => o = ⌜false⌝ ; (s)

Proof step 4 from proof ls-null (previous step) (back to tree)

User applied rule expand left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ rP)(h) ↔ (∃ r. rP(r)(h)); (s) ⊦ ¬ false(h); (s)