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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  h0 ∪ h2 = h1 ∪ h3, h0 ⊥ h2, h1 ⊥ h3, (hP0 * hP1 * true)(h0 ∪ h2), hP0(h0), hP1(h1), precise(hP1), precise(hP0), ¬ h0 ⊥ h1 ⊦

Rule argument

Additional information

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  h0 ∪ h2 = h1 ∪ h3, h0 ⊥ h2, h1 ⊥ h3, hP0(h0), (hP0 * hP1 * true)(h1 ∪ h3), hP1(h1), precise(hP0), precise(hP1), ¬ h0 ⊥ h1 ⊦

Rule argument

Additional information

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

User applied rule apply lemma

And obtained the following new goals

• Proof step 4:

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

• Proof step 5:

  h0 ∪ h2 = h1 ∪ h3, h0 ⊥ h2, h1 ⊥ h3, (hP0 * hP1 * true)(h1 ∪ h3), hP0(h0), hP1(h1), precise(hP1), precise(hP0), ¬ h0 ⊥ h1, (hP0 * hP1 * true)(h1 ∪ h3) ↔ (∃ h0, h2. h0 ⊥ h2 ∧ h1 ∪ h3 = h0 ∪ h2 ∧ hP0(h0) ∧ (hP1 * true)(h2)) ⊦

Rule argument

Additional information

Used terms: hP0, hP1 * true, h1 ∪ h3

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, (hP0 * hP1 * true)(h ∪ h4), hP0(h), hP1(h1), (hP1 * true)(h4), precise(hP0), precise(hP1), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used simplifier rules: true-elim: ⊦ (hP0 * hP)(h) → (true * hP)(h); sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; sep-com: ⊦ hP0 * hP1 = hP1 * hP0; precise-same: precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ hP0 * hP1 = hP1 * hP0

• Proof step 8:

  h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, hP0(h), (hP1 * true)(h4), hP1(h1), ((hP1 * true) * hP0)(h ∪ h4), precise(hP1), precise(hP0), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used terms: hP1 * true, hP0

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

Lemma sep-com

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2

• Proof step 10:

  h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, hP0(h), (hP1 * true * hP0)(h ∪ h4), hP1(h1), (hP1 * true)(h4), precise(hP0), precise(hP1), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used terms: hP0, true, hP1

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

Lemma sep-ass

Proof step 10 from proof precise-disj (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:

  h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, hP0(h), (hP1 * true)(h4), hP1(h1), precise(hP1), precise(hP0), ¬ h ⊥ h1, ∃ h0, h1. h0 ⊥ h1 ∧ h ∪ h4 = h0 ∪ h1 ∧ hP1(h0) ∧ (true * hP0)(h1) ⊦

Rule argument

Additional information

Used terms: h ∪ h4, true * hP0, hP1

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h ∪ h4 = h0 ∪ h2, h ⊥ h4, h0 ⊥ h2, hP0(h), hP1(h0), (hP1 * true)(h4), precise(hP0), precise(hP1), (true * hP0)(h2), ¬ h ⊥ h0 ⊦

Rule argument

Additional information

Used simplifier rules: precise-same: precise(hP) ⊦ hP(h0) ∧ hP(h1) ∧ h0 ⊥ h ∧ h1 ⊥ h2 → (h0 ∪ h = h1 ∪ h2 ↔ h0 = h1 ∧ h = h2);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ hP0 * hP1 = hP1 * hP0

• Proof step 15:

  h ∪ h4 = h0 ∪ h2, h ⊥ h4, h0 ⊥ h2, (hP0 * true)(h2), hP0(h), (hP1 * true)(h4), hP1(h0), precise(hP1), precise(hP0), ¬ h ⊥ h0 ⊦

Rule argument

Additional information

Used terms: hP0, true

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

Lemma sep-com

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

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

• Proof step 17:

  h ∪ h4 = h0 ∪ h2, h ⊥ h4, h0 ⊥ h2, hP0(h), hP1(h0), (hP1 * true)(h4), precise(hP0), precise(hP1), ¬ h ⊥ h0, ∃ h0, h1. h0 ⊥ h1 ∧ h2 = h0 ∪ h1 ∧ hP0(h0) ∧ true(h1) ⊦

Rule argument

Additional information

Used terms: h2, true, hP0

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  h ∪ h4 = h0 ∪ h1 ∪ h3, h ⊥ h4, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, hP0(h1), hP0(h), (hP1 * true)(h4), hP1(h0), precise(hP1), precise(hP0), ¬ h ⊥ h0 ⊦

Rule argument

Additional information

Used simplifier rules: trueH: ⊦ true(h);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

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

• Proof step 20:

  h ∪ h4 = h0 ∪ h1 ∪ h3, h ⊥ h4, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, hP0(h), hP0(h1), hP1(h0), precise(hP0), precise(hP1), ¬ h ⊥ h0, ∃ h0, h1. h0 ⊥ h1 ∧ h4 = h0 ∪ h1 ∧ hP1(h0) ∧ true(h1) ⊦

Rule argument

Additional information

Used terms: h4, true, hP1

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  h ∪ h2 ∪ h5 = h0 ∪ h1 ∪ h3, h ⊥ h2 ∪ h5, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, h2 ⊥ h5, hP0(h1), hP0(h), hP1(h2), hP1(h0), precise(hP1), precise(hP0), ¬ h ⊥ h0 ⊦

Rule argument

Additional information

Used simplifier rules: trueH: ⊦ true(h);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

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

• Proof step 23:

  h ∪ h2 ∪ h5 = h0 ∪ h1 ∪ h3, h ⊥ h2 ∪ h5, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, h2 ⊥ h5, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), ¬ h ⊥ h0, ∀ h, h0, h1, h2. hP0(h) ∧ hP0(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0 ⊦

Rule argument

Additional information

Used terms: hP0

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

Lemma precise

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

User applied rule all left

And obtained the following new goals

• Proof step 24:

  h ∪ h2 ∪ h5 = h0 ∪ h1 ∪ h3, h ⊥ h2 ∪ h5, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, h2 ⊥ h5, hP0(h1), hP0(h), hP1(h2), hP1(h0), precise(hP1), ¬ h ⊥ h0, ∀ h, h0, h1, h2. hP0(h) ∧ hP0(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0, hP0(h) ∧ hP0(h1) ∧ h ⊥ h2 ∪ h5 ∧ h1 ⊥ h0 ∪ h3 ∧ h ∪ h2 ∪ h5 = h1 ∪ h0 ∪ h3 → h = h1 ⊦

Rule argument

Additional information

Used terms: h, h1, h2 ∪ h5, h0 ∪ h3

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  h ∪ h2 ∪ h5 = h0 ∪ h1 ∪ h3, h ⊥ h2 ∪ h5, h0 ⊥ h1 ∪ h3, h1 ⊥ h3, h2 ⊥ h5, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), ¬ h ⊥ h0, ∀ h, h0, h1, h2. hP0(h) ∧ hP0(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0, h1 ⊥ h0 ∪ h3 ∧ h0 ∪ h1 ∪ h3 = h1 ∪ h0 ∪ h3 → h = h1 ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

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

• Proof step 27:

  h ∪ h2 ∪ h5 = h0 ∪ h1 ∪ h3, h ⊥ h2 ∪ h5, h0 ⊥ h1 ∪ h3, h0 ⊥ h3, h1 ⊥ h3, h1 ⊥ h0, h2 ⊥ h5, h3 ⊥ h1, hP0(h1), hP0(h), hP1(h2), hP1(h0), precise(hP1), ¬ h ⊥ h0, ∀ h, h0, h1, h2. hP0(h) ∧ hP0(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0, h1 ⊥ h0 ∪ h3 ∧ h0 ∪ h1 ∪ h3 = (h1 ∪ h0) ∪ h3 → h = h1 ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws) h0 ⊥ h1, h1 ⊥ h, h ⊥ h0 ⊦ h ∪ h0 ∪ h1 = h0 ∪ h ∪ h1 ↔ true; (ss) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h1, h0, h3

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

A spec-lemma

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Rule argument

Additional information

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

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

A spec-lemma