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

Rule argument

Additional information

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

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, hP0(h1), hP0(h), hP1(h0), precise(hP0), precise(hP1), h0 ⊥ h3, h0 ⊥ h1, h ⊥ h4, h1 ⊥ h3, ¬ 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, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), precise(hP0), h0 ⊥ h1, h0 ⊥ h3, h2 ⊥ h5, h1 ⊥ h3, h ⊥ h2, h ⊥ h5, ¬ h ⊥ h0 ⊦

Rule argument

Additional information

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

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

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 ≠ h1, h0 ∪ h1 ∪ h3 ≠ h1 ∪ h0 ∪ h3, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), h0 ⊥ h1, h0 ⊥ h3, h2 ⊥ h5, h1 ⊥ h3, h ⊥ h2, h ⊥ h5, ¬ 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 simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) ⊦ h1 ⊥ h → (h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h); (ws)

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 ≠ h1, (h0 ∪ h1) ∪ h3 ≠ h1 ∪ h0 ∪ h3, hP0(h1), hP0(h), hP1(h2), hP1(h0), precise(hP1), h ⊥ h5, h ⊥ h2, h2 ⊥ h5, h0 ⊥ h3, h0 ⊥ h1, h1 ⊥ h3, h3 ⊥ h0, ¬ 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 simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h3, h1, h0

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  h ∪ h2 ∪ h5 = (h0 ∪ h1) ∪ h3, h ≠ h1, (h0 ∪ h1) ∪ h3 ≠ h1 ∪ h0 ∪ h3, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), h1 ⊥ h3, h0 ⊥ h1, h0 ⊥ h3, h2 ⊥ h5, h ⊥ h2, h ⊥ h5, ¬ 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 simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 29:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 30:

  h ∪ h2 ∪ h5 = (h1 ∪ h0) ∪ h3, h ≠ h1, (h1 ∪ h0) ∪ h3 ≠ h1 ∪ h0 ∪ h3, hP0(h1), hP0(h), hP1(h2), hP1(h0), precise(hP1), h ⊥ h5, h ⊥ h2, h2 ⊥ h5, h0 ⊥ h3, h0 ⊥ h1, h1 ⊥ h3, h1 ⊥ h0, ¬ 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 simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h1, h0

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 31:

  h ∪ h2 ∪ h5 = (h1 ∪ h0) ∪ h3, h ≠ h1, (h1 ∪ h0) ∪ h3 ≠ h1 ∪ h0 ∪ h3, hP0(h), hP0(h1), hP1(h0), hP1(h2), precise(hP1), h1 ⊥ h3, h0 ⊥ h1, h0 ⊥ h3, h2 ⊥ h5, h ⊥ h2, h ⊥ h5, ¬ 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 simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

  ⊦ h0 ⊥ h1 ∧ h1 ⊥ h ∧ h ⊥ h0 → (h0 ∪ h1) ∪ h = h0 ∪ h1 ∪ h

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (s) Used terms: h3, h0, h1

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

A spec-lemma