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

Used simplifier rules: true: ⊦ true = ⌜true⌝ ;

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

Rule argument

Additional information

Used terms: hP0, ⌜true⌝ * hP1, 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), precise(hP0), precise(hP1), (⌜true⌝ * hP1)(h4), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used simplifier rules: 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)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 8:

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

Rule argument

Additional information

Used simplifier rules: sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; sep-com: ⊦ hP0 * hP1 = hP1 * hP0; Used terms: h ∪ h4, hP0 * ⌜true⌝ , hP1

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  h ∪ h4 = h0 ∪ h2, h1 ∪ h3 = h0 ∪ h2, h ⊥ h4, h0 ⊥ h2, h1 ⊥ h3, (hP0 * ⌜true⌝ )(h0), hP0(h), hP1(h2), hP1(h1), precise(hP0), precise(hP1), (⌜true⌝ * hP1)(h4), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  h ∪ h4 = h0 ∪ h2, h1 ∪ h3 = h0 ∪ h2, h ⊥ h4, h0 ⊥ h2, h1 ⊥ h3, hP0(h), hP1(h1), hP1(h2), precise(hP1), precise(hP0), (⌜true⌝ * hP1)(h4), ¬ h ⊥ h1, ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ hP0(h1) ∧ ⌜true⌝ (h2) ⊦

Rule argument

Additional information

Used terms: h0, hP0, ⌜true⌝

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  h ∪ h4 = (h5 ∪ h6) ∪ h2, h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h6, h5 ⊥ h6, hP0(h5), hP0(h), hP1(h2), hP1(h1), precise(hP0), precise(hP1), (⌜true⌝ * hP1)(h4), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used simplifier rules: prop: ⊦ ⌜bv⌝ (h) ↔ bv; h1 ⊥ h2 ⊦ h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

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

• Proof step 14:

  h ∪ h4 = (h5 ∪ h6) ∪ h2, h1 ∪ h3 = h ∪ h4, h ⊥ h4, h1 ⊥ h3, h2 ⊥ h6, h2 ⊥ h5, h5 ⊥ h6, hP0(h), hP0(h5), hP1(h1), hP1(h2), precise(hP1), precise(hP0), ¬ h ⊥ h1, ∃ h0, h1. h0 ⊥ h1 ∧ h4 = h0 ∪ h1 ∧ hP1(h0) ∧ ⌜true⌝ (h1) ⊦

Rule argument

Additional information

Used terms: h4, hP1, ⌜true⌝

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

Lemma sep

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  h ∪ h0 ∪ h7 = h1 ∪ h3, h1 ∪ h3 = (h5 ∪ h6) ∪ h2, h ⊥ h0, h ⊥ h7, h0 ⊥ h7, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h6, h5 ⊥ h6, hP0(h5), hP0(h), hP1(h0), hP1(h2), hP1(h1), precise(hP0), precise(hP1), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

Used simplifier rules: prop: ⊦ ⌜bv⌝ (h) ↔ bv; h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws)

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:

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

• Proof step 17:

  h ∪ h0 ∪ h7 = h1 ∪ h3, h1 ∪ h3 = (h5 ∪ h6) ∪ h2, h ⊥ h7, h ⊥ h0, h0 ⊥ h7, h1 ⊥ h3, h2 ⊥ h6, h2 ⊥ h5, h5 ⊥ h6, hP0(h), hP0(h5), hP1(h1), hP1(h2), hP1(h0), precise(hP0), ¬ h ⊥ h1, ∀ h, h0, h1, h2. hP1(h) ∧ hP1(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0 ⊦

Rule argument

Additional information

Used terms: hP1

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

Lemma precise

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

User applied rule all left

And obtained the following new goals

• Proof step 18:

  h ∪ h0 ∪ h7 = h1 ∪ h3, h1 ∪ h3 = (h5 ∪ h6) ∪ h2, h ⊥ h0, h ⊥ h7, h0 ⊥ h7, h1 ⊥ h3, h2 ⊥ h5, h2 ⊥ h6, h5 ⊥ h6, hP0(h5), hP0(h), hP1(h0), hP1(h2), hP1(h1), precise(hP0), ¬ h ⊥ h1, hP1(h1) ∧ hP1(h0) ∧ h1 ⊥ h3 ∧ h0 ⊥ h ∪ h7 ∧ h1 ∪ h3 = h0 ∪ h ∪ h7 → h1 = h0 ⊦

Rule argument

Additional information

Used terms: h1, h0, h3, h ∪ h7

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  h ∪ h0 ∪ h7 = h1 ∪ h3, h1 ∪ h3 = (h5 ∪ h6) ∪ h2, h1 ∪ h3 ≠ h0 ∪ h ∪ h7, h ⊥ h7, h ⊥ h0, h0 ⊥ h7, h1 ⊥ h3, h2 ⊥ h6, h2 ⊥ h5, h5 ⊥ h6, hP0(h), hP0(h5), hP1(h1), hP1(h2), hP1(h0), precise(hP0), ¬ h ⊥ h1 ⊦

Rule argument

Additional information

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

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

User applied rule subst equation

And closed the goal!

Rule argument

Additional information

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