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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬ precise(ls(r, x)) ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ precise(hP) ↔ (∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0)

• Proof step 4:

  ∃ h, h0, h1, h2. ¬(ls(r, x)(h) ∧ ls(r, x)(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0) ⊦

Rule argument

Additional information

Used terms: ls(r, x)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, ls(r, x)(h), ls(r, x)(h0) ⊦

Rule argument

Additional information

Proof step 5 from proof precise-ls (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, ls(r, a + x0)(h0), ls(r, a + x0)(h), ∀ h, h1, h0, h2, r. ¬(h ∪ h1 = h0 ∪ h2 ∧ h ≠ h0 ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ ls(r, x0)(h) ∧ ls(r, x0)(h0)) ⊦

Rule argument

Additional information

Used simplifier rules: ls-base: ⊦ ls(r, []) = (emp ∧ ⌜r = null⌝ ); List[A] freely generated by [], +; ⊦ (hP0 ∧ hP1)(h) ↔ hP0(h) ∧ hP1(h); (s) ⊦ ⌜bv⌝ (h) ↔ bv; (s) ⊦ emp(h) ↔ h = ∅; (s) ⊦ ⌜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 ⊦ precise(emp); (s) precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2); (s)

Proof step 6 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, (r => a × r0 * ls(r0, x0))(h), (r => a × r1 * ls(r1, x0))(h0), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: ls-rec: ⊦ ls(r, a ' + x) = _∃(λ r0. r => a × r0 * ls(r0, x)); ⊦ (_∃ bP)(h) ↔ (∃ b. bP(b)(h)); (s) a: ⊦ (x + y) + z = x + y + z; from specification list ⊦ a + x = a ' + x; (s)

Proof step 7 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 9:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, (r => a × r1 * ls(r1, x0))(h0), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ (r => a × r0)(h0) ∧ ls(r0, x0)(h1), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used terms: h, ls(r0, x0), r => a × r0

Proof step 8 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  (h3 ∪ h4) ∪ h1 = h0 ∪ h2, h3 ∪ h4 ≠ h0, h0 ⊥ h2, h1 ⊥ h3, h1 ⊥ h4, h3 ⊥ h4, ls(r0, x0)(h4), (r => a × r0)(h3), (r => a × r1 * ls(r1, x0))(h0), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s)

Proof step 10 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

  ⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 12:

  (h3 ∪ h4) ∪ h1 = h0 ∪ h2, h3 ∪ h4 ≠ h0, h0 ⊥ h2, h1 ⊥ h4, h1 ⊥ h3, h3 ⊥ h4, ls(r0, x0)(h4), (r => a × r0)(h3), ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ (r => a × r1)(h1) ∧ ls(r1, x0)(h2), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used terms: h0, ls(r1, x0), r => a × r1

Proof step 11 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  (h3 ∪ h4) ∪ h1 = (h ∪ h5) ∪ h2, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h3, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h5), ls(r0, x0)(h4), (r => a × r1)(h), (r => a × r0)(h3), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s)

Proof step 13 from proof precise-ls (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 14:

  (h3 ∪ h4) ∪ h1 = (h ∪ h5) ∪ h2, h3 = h, r0 = r1, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, ls(r0, x0)(h4), ls(r1, x0)(h5), (r => a × r0)(h3), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

• Proof step 21:

  (h3 ∪ h4) ∪ h1 = (h ∪ h5) ∪ h2, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, ls(r0, x0)(h4), ls(r1, x0)(h5), (r => a × r0)(h3), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used formulas: h3 = h ∧ r0 = r1

Proof step 14 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  (h ∪ h4) ∪ h1 = (h ∪ h5) ∪ h2, h4 ≠ h5, h ⊥ h4, h ⊥ h5, h1 ⊥ h, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, ls(r1, x0)(h5), ls(r1, x0)(h4), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2); (s) ⊦ precise(r => o); (s)

Proof step 15 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  h0 ⊥ h1, h1 ⊥ h, h ⊥ h0 ⊦ (h0 ∪ h1) ∪ h = h0 ∪ h1 ∪ h

• Proof step 17:

  h ∪ h4 ∪ h1 = (h ∪ h5) ∪ h2, h4 ≠ h5, h ⊥ h5, h ⊥ h4, h1 ⊥ h4, h1 ⊥ h, h2 ⊥ h5, h2 ⊥ h, h4 ⊥ h1, ls(r1, x0)(h4), ls(r1, x0)(h5), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h, h4, h1

Proof step 16 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

  h0 ⊥ h1, h1 ⊥ h, h ⊥ h0 ⊦ (h0 ∪ h1) ∪ h = h0 ∪ h1 ∪ h

• Proof step 19:

  h ∪ h4 ∪ h1 = h ∪ h5 ∪ h2, h4 ≠ h5, h ⊥ h4, h ⊥ h5, h1 ⊥ h, h1 ⊥ h4, h2 ⊥ h5, h2 ⊥ h, h4 ⊥ h1, h5 ⊥ h2, ls(r1, x0)(h5), ls(r1, x0)(h4), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h, h5, h2

Proof step 18 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  h4 ∪ h1 = h5 ∪ h2, h4 ≠ h5, h ⊥ h5, h ⊥ h4, h1 ⊥ h4, h1 ⊥ h, h2 ⊥ h, h2 ⊥ h5, h4 ⊥ h1, h5 ⊥ h2, ls(r1, x0)(h4), ls(r1, x0)(h5), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5) ⊦

Rule argument

Additional information

Used simplifier rules: h ⊥ h0, h ⊥ h1 ⊦ h ∪ h0 = h ∪ h1 ↔ h0 = h1; (s) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws)

Proof step 20 from proof precise-ls (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h4, h1, h5, h2, r1

Proof step 21 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 23:

  h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h3, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h5), ls(r0, x0)(h4), (r => a × r1)(h), (r => a × r0)(h3), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used terms: (h3 ∪ h4) ∪ h1, (h ∪ h5) ∪ h2

Proof step 22 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof precise-ls (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 24:

  r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, ls(r0, x0)(h4), ls(r1, x0)(h5), (r => a × r0)(h3), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)), r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used terms: r

Proof step 24 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h3, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h5), ls(r0, x0)(h4), (r => a × r1)(h), (r => a × r0)(h3), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), r ∈ (h ∪ h5) ∪ h2 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Proof step 25 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))

• Proof step 27:

  r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2, h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, h3 ∪ h4 ⊥ h1, ls(r0, x0)(h4), ls(r1, x0)(h5), (r => a × r0)(h3), (r => a × r1)(h), r ∈ (h3 ∪ h4) ∪ h1, ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)), r ∈ (h ∪ h5) ∪ h2 → (r ∈ h3 ∪ h4 ⊃ (h3 ∪ h4)(r);h1(r)) = ((h ∪ h5) ∪ h2)(r) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ (r => o)(h) → r ∈ h; (s) h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s) Used terms: r, h3 ∪ h4, h1

Proof step 26 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  a × r0 = ((h ∪ h5) ∪ h2)(r), h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h1 ⊥ h3, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h5), ls(r0, x0)(h4), (r => a × r1)(h), (r => a × r0)(h3), r ∈ (h ∪ h5) ∪ h2, ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used simplifier rules: ⊦ (r => o0)(h) → h(r) = o0; (s) ⊦ (r => o)(h) → r ∈ h; (s) r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s)

Proof step 28 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))

• Proof step 30:

  a × r0 = (r ∈ h ∪ h5 ⊃ (h ∪ h5)(r);h2(r)), h3 ∪ h4 ≠ h ∪ h5, h ⊥ h5, h ∪ h5 ⊥ h2, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, ls(r0, x0)(h4), ls(r1, x0)(h5), (r => a × r0)(h3), (r => a × r1)(h), r ∈ (h ∪ h5) ∪ h2, ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦ h3 = h ∧ r0 = r1

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ (r => o)(h) → r ∈ h; (s) h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s) Used terms: r, h ∪ h5, h2

Proof step 29 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 30 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  h3 ≠ h, h ⊥ h5, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h, h3 ⊥ h4, ls(r1, x0)(h4), ls(r1, x0)(h5), (r => a × r1)(h3), (r => a × r1)(h), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦

Rule argument

Additional information

Used simplifier rules: precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2); (s) ⊦ precise(r => o); (s) h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s) ⊦ (r => o)(h) → r ∈ h; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ a × r = a0 × r0 ↔ a = a0 ∧ r = r0; (ws) ⊦ (r => o0)(h) → h(r) = o0; (s) r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s)

Proof step 31 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

• Proof step 33:

  h = (∅ ++ r)(r := a × r1), h3 ≠ h, r ≠ null, h ⊥ h5, h1 ⊥ h3, h1 ⊥ h4, h2 ⊥ h, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h5), ls(r1, x0)(h4), (r => a × r1)(h3), ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r. (r ∈ (h3 ∪ h4) ∪ h1 ↔ r ∈ (h ∪ h5) ∪ h2) ∧ (r ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r) = ((h ∪ h5) ∪ h2)(r)) ⊦

Rule argument

Additional information

Used terms: r, a × r1, h

Proof step 32 from proof precise-ls (previous step) (back to tree)

A spec-lemma

Proof step 33 from proof precise-ls (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  h3 ≠ (∅ ++ r)(r := a × r1), r ≠ null, h1 ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h3 ⊥ h4, ls(r1, x0)(h4), ls(r1, x0)(h5), (r => a × r1)(h3), ¬ r ∈ h2, ¬ r ∈ h5, ∀ h3, h4, h5, h6, r2. h3 ∪ h4 ≠ h5 ∪ h6 ∨ h3 = h5 ∨ ¬ h3 ⊥ h4 ∨ ¬ h5 ⊥ h6 ∨ ¬ ls(r2, x0)(h3) ∨ ¬ ls(r2, x0)(h5), ∀ r0. (r0 ∈ (h3 ∪ h4) ∪ h1 ↔ r0 ∈ ((∅ ++ r)(r := a × r1) ∪ h5) ∪ h2) ∧ (r0 ∈ (h3 ∪ h4) ∪ h1 → ((h3 ∪ h4) ∪ h1)(r0) = (((∅ ++ r)(r := a × r1) ∪ h5) ∪ h2)(r0)) ⊦

Rule argument

Additional information

Used simplifier rules: r ≠ null ⊦ (∅ ++ r)(r := o) ⊥ h ↔ ¬ r ∈ h; (s) ⊦ h ⊥ ∅; (s) r ≠ null, ¬ r ∈ h1 ⊦ h0 ⊥ (h1 ++ r)(r := o) ↔ h0 ⊥ h1 ∧ ¬ r ∈ h0; (s) ⊦ ¬ r ∈ ∅; (s)

Proof step 34 from proof precise-ls (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

Rule argument

Additional information

Used terms: r, a × r1, h3

Proof step 35 from proof precise-ls (previous step) (back to tree)

A spec-lemma