Proof step 1 from proof lls-one (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  lls(r0, r1, a ') ≠ r0 => a × r1 ⊦

Rule argument

Additional information

Proof step 2 from proof lls-one (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ a ' = a + []

• Proof step 4:

  lls(r0, r1, a + []) ≠ r0 => a × r1 ⊦

Rule argument

Additional information

Used terms: a

Proof step 3 from proof lls-one (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof lls-one (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

  ⊦ a + x = a ' + x

• Proof step 6:

  lls(r0, r1, a ' + []) ≠ r0 => a × r1 ⊦

Rule argument

Additional information

Used terms: [], a

Proof step 5 from proof lls-one (previous step) (back to tree)

A spec-lemma

Proof step 6 from proof lls-one (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ lls(r, r1, a ' + x) = _∃(λ r0. r => a × r0 * lls(r0, r1, x))

• Proof step 8:

  _∃(λ r2. r0 => a × r2 * lls(r2, r1, [])) ≠ r0 => a × r1 ⊦

Rule argument

Additional information

Used terms: [], a, r1, r0

Proof step 7 from proof lls-one (previous step) (back to tree)

Lemma lls-rec

Proof step 8 from proof lls-one (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  _∃(λ r2. ⌜r2 = r1⌝ ∧ r0 => a × r2) ≠ r0 => a × r1 ⊦

Rule argument

Additional information

Used simplifier rules: lls-base: ⊦ lls(r0, r1, []) = (emp ∧ ⌜r0 = r1⌝ ); sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep ⊦ emp * hP = hP; (s) ⊦ (⌜bv⌝ ∧ hP0) * hP1 = (⌜bv⌝ ∧ hP0 * hP1); (s)

Proof step 9 from proof lls-one (previous step) (back to tree)

User applied rule expand left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜false⌝ = false; (s) ⊦ ¬ false(h); (s) ⊦ (_∃ rP)(h) ↔ (∃ r. rP(r)(h)); (s) ⊦ ((hP0 ∧ hP1))(h) ↔ hP0(h) ∧ hP1(h); (s) ⊦ ⌜bv⌝ (h) ↔ bv; (s) and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep