Proof step 1 from proof introduce-sharing (conclusion) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 2:

  (hP0 ⊗ hP1)(h) ⊦ ∃ hP. ((hP -* hP0) * (hP -* hP1) * hP)(h)

• Proof step 23:

  ¬ (hP0 ⊗ hP1)(h), ∃ hP. ((hP -* hP0) * (hP -* hP1) * hP)(h) ⊦

Rule argument

Additional information

Proof step 2 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ (hP0 ⊗ hP1)(h) ↔ (∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP0(h0 ∪ h2) ∧ hP1(h1 ∪ h2))

• Proof step 4:

  ∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP0(h0 ∪ h2) ∧ hP1(h1 ∪ h2) ⊦ ∃ hP. ((hP -* hP0) * (hP -* hP1) * hP)(h)

Rule argument

Additional information

Used terms: h, hP1, hP0

Proof step 3 from proof introduce-sharing (previous step) (back to tree)

Lemma sharing

Proof step 4 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2) ⊦ ∃ hP. ((hP -* hP0) * (hP -* hP1) * hP)(h0 ∪ h1 ∪ h2)

Rule argument

Additional information

Proof step 5 from proof introduce-sharing (previous step) (back to tree)

User applied rule exists right

And obtained the following new goals

• Proof step 6:

  h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2) ⊦

Rule argument

Additional information

Used terms: {h2}

Proof step 6 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)

• Proof step 8:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2) ⊦ ({h2} -* hP0)(h0) ∧ (({h2} -* hP1) * {h2} )(h1 ∪ h2) ∧ h0 ⊥ h1 ∪ h2

Rule argument

Additional information

Used terms: h1 ∪ h2, h0, ({h2} -* hP1) * {h2} , {h2} -* hP0

Proof step 7 from proof introduce-sharing (previous step) (back to tree)

Lemma split-union

Proof step 8 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2) ⊦ ({h2} -* hP0)(h0) ∧ (({h2} -* hP1) * {h2} )(h1 ∪ h2)

Rule argument

Additional information

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

Proof step 9 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

  ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)

• Proof step 11:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2) ⊦ ({h2} -* hP0)(h0) ∧ (({h2} -* hP1) * {h2} )(h1 ∪ h2), ({h2} -* hP1)(h1) ∧ {h2} (h2) ∧ h1 ⊥ h2

• Proof step 17:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ({h2} -* hP1)(h1), {h2} (h2), hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2) ⊦ ({h2} -* hP0)(h0) ∧ true

Rule argument

Additional information

Used terms: h2, h1, {h2} , {h2} -* hP1

Proof step 10 from proof introduce-sharing (previous step) (back to tree)

Lemma split-union

Proof step 11 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2), ¬ ({h2} -* hP1)(h1) ⊦ ({h2} -* hP0)(h0) ∧ (({h2} -* hP1) * {h2} )(h1 ∪ h2)

Rule argument

Additional information

Used simplifier rules: this: ⊦ {h} (h0) ↔ h = h0;

Proof step 12 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 13:

  ⊦ (hP0 -* hP1)(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 14:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2), ∃ h0. ¬(h0 ⊥ h1 ∧ {h2} (h0) → hP1(h0 ∪ h1)) ⊦ ({h2} -* hP0)(h0) ∧ (({h2} -* hP1) * {h2} )(h1 ∪ h2)

Rule argument

Additional information

Used terms: h1, hP1, {h2}

Proof step 13 from proof introduce-sharing (previous step) (back to tree)

Lemma wand

Proof step 14 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  h ⊥ h1, h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h, hP0(h0 ∪ h), hP1(h1 ∪ h), ¬ (({h} -* hP0) * ({h} -* hP1) * {h} )(h0 ∪ h1 ∪ h), ¬ hP1(h ∪ h1) ⊦ ({h} -* hP0)(h0) ∧ (({h} -* hP1) * {h} )(h1 ∪ h)

Rule argument

Additional information

Used simplifier rules: this: ⊦ {h} (h0) ↔ h = h0;

Proof step 15 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Rule argument

Additional information

Used terms: h1, h

Proof step 16 from proof introduce-sharing (previous step) (back to tree)

A spec-lemma

Proof step 17 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

  h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, ({h2} -* hP1)(h1), hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2), ¬ ({h2} -* hP0)(h0) ⊦

Rule argument

Additional information

Used simplifier rules: this: ⊦ {h} (h0) ↔ h = h0;

Proof step 18 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 19:

  ⊦ (hP0 -* hP1)(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 20:

  h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, ({h2} -* hP1)(h1), hP0(h0 ∪ h2), hP1(h1 ∪ h2), ¬ (({h2} -* hP0) * ({h2} -* hP1) * {h2} )(h0 ∪ h1 ∪ h2), ∃ h1. ¬(h1 ⊥ h0 ∧ {h2} (h1) → hP0(h1 ∪ h0)) ⊦

Rule argument

Additional information

Used terms: h0, hP0, {h2}

Proof step 19 from proof introduce-sharing (previous step) (back to tree)

Lemma wand

Proof step 20 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 21:

  h ⊥ h0, ({h} -* hP1)(h1), h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h, hP0(h0 ∪ h), hP1(h1 ∪ h), ¬ (({h} -* hP0) * ({h} -* hP1) * {h} )(h0 ∪ h1 ∪ h), ¬ hP0(h ∪ h0) ⊦

Rule argument

Additional information

Used simplifier rules: this: ⊦ {h} (h0) ↔ h = h0;

Proof step 21 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 22:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Rule argument

Additional information

Used terms: h0, h

Proof step 22 from proof introduce-sharing (previous step) (back to tree)

A spec-lemma

Proof step 23 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 24:

  ((hP -* hP0) * (hP -* hP1) * hP)(h), ¬ (hP0 ⊗ hP1)(h) ⊦

Rule argument

Additional information

Proof step 24 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 25:

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

• Proof step 26:

  ¬ (hP0 ⊗ hP1)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ (hP -* hP0)(h0) ∧ ((hP -* hP1) * hP)(h1) ⊦

Rule argument

Additional information

Used terms: h, (hP -* hP1) * hP, hP -* hP0

Proof step 25 from proof introduce-sharing (previous step) (back to tree)

Lemma sep

Proof step 26 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 27:

  h0 ⊥ h1, (hP -* hP0)(h0), ((hP -* hP1) * hP)(h1), ¬ (hP0 ⊗ hP1)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 27 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 28:

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

• Proof step 29:

  h0 ⊥ h1, (hP -* hP0)(h0), ¬ (hP0 ⊗ hP1)(h0 ∪ h1), ∃ h0, h2. h0 ⊥ h2 ∧ h1 = h0 ∪ h2 ∧ (hP -* hP1)(h0) ∧ hP(h2) ⊦

Rule argument

Additional information

Used terms: h1, hP, hP -* hP1

Proof step 28 from proof introduce-sharing (previous step) (back to tree)

Lemma sep

Proof step 29 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, (hP -* hP1)(h), hP(h2), (hP -* hP0)(h0), ¬ (hP0 ⊗ hP1)(h0 ∪ h ∪ h2) ⊦

Rule argument

Additional information

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

Proof step 30 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 31:

  ⊦ (hP0 ⊗ hP1)(h) ↔ (∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ hP0(h0 ∪ h2) ∧ hP1(h1 ∪ h2))

• Proof step 32:

  h ⊥ h2, h0 ⊥ h2, h0 ⊥ h, (hP -* hP0)(h0), hP(h2), (hP -* hP1)(h), ∀ h1, h3, h4. ¬(h1 ⊥ h4 ∧ h1 ⊥ h3 ∧ h3 ⊥ h4 ∧ h0 ∪ h ∪ h2 = h1 ∪ h3 ∪ h4 ∧ hP0(h1 ∪ h4) ∧ hP1(h3 ∪ h4)) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h ∪ h2, hP1, hP0

Proof step 31 from proof introduce-sharing (previous step) (back to tree)

Lemma sharing

Proof step 32 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 33:

  ⊦ (hP0 -* hP1)(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 34:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, (hP -* hP1)(h), hP(h2), ∀ h1, h3, h4. ¬(h1 ⊥ h4 ∧ h1 ⊥ h3 ∧ h3 ⊥ h4 ∧ h0 ∪ h ∪ h2 = h1 ∪ h3 ∪ h4 ∧ hP0(h1 ∪ h4) ∧ hP1(h3 ∪ h4)), ∀ h1. h1 ⊥ h0 ∧ hP(h1) → hP0(h1 ∪ h0) ⊦

Rule argument

Additional information

Used terms: h0, hP0, hP

Proof step 33 from proof introduce-sharing (previous step) (back to tree)

Lemma wand

Proof step 34 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ (hP0 -* hP1)(h) ↔ (∀ h1. h1 ⊥ h ∧ hP0(h1) → hP1(h1 ∪ h))

• Proof step 36:

  h ⊥ h2, h0 ⊥ h2, h0 ⊥ h, hP(h2), ∀ h1. h1 ⊥ h0 ∧ hP(h1) → hP0(h1 ∪ h0), ∀ h1, h3, h4. ¬(h1 ⊥ h4 ∧ h1 ⊥ h3 ∧ h3 ⊥ h4 ∧ h0 ∪ h ∪ h2 = h1 ∪ h3 ∪ h4 ∧ hP0(h1 ∪ h4) ∧ hP1(h3 ∪ h4)), ∀ h1. h1 ⊥ h ∧ hP(h1) → hP1(h1 ∪ h) ⊦

Rule argument

Additional information

Used terms: h, hP1, hP

Proof step 35 from proof introduce-sharing (previous step) (back to tree)

Lemma wand

Proof step 36 from proof introduce-sharing (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 37:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, hP(h2), ∀ h1. h1 ⊥ h ∧ hP(h1) → hP1(h1 ∪ h), ∀ h1, h3, h4. ¬(h1 ⊥ h4 ∧ h1 ⊥ h3 ∧ h3 ⊥ h4 ∧ h0 ∪ h ∪ h2 = h1 ∪ h3 ∪ h4 ∧ hP0(h1 ∪ h4) ∧ hP1(h3 ∪ h4)), h2 ⊥ h0 ∧ hP(h2) → hP0(h2 ∪ h0) ⊦

Rule argument

Additional information

Used terms: h2

Proof step 37 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 38:

  h ⊥ h2, h0 ⊥ h2, h0 ⊥ h, hP(h2), hP0(h2 ∪ h0), ∀ h1. h1 ⊥ h ∧ hP(h1) → hP1(h1 ∪ h), ∀ h1, h3, h4. ¬ h1 ⊥ h4 ∨ ¬ h1 ⊥ h3 ∨ ¬ h3 ⊥ h4 ∨ h0 ∪ h ∪ h2 ≠ h1 ∪ h3 ∪ h4 ∨ ¬ hP0(h1 ∪ h4) ∨ ¬ hP1(h3 ∪ h4) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss)

Proof step 38 from proof introduce-sharing (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 39:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, hP(h2), hP0(h2 ∪ h0), ∀ h1, h3, h4. ¬ h1 ⊥ h4 ∨ ¬ h1 ⊥ h3 ∨ ¬ h3 ⊥ h4 ∨ h0 ∪ h ∪ h2 ≠ h1 ∪ h3 ∪ h4 ∨ ¬ hP0(h1 ∪ h4) ∨ ¬ hP1(h3 ∪ h4), h2 ⊥ h ∧ hP(h2) → hP1(h2 ∪ h) ⊦

Rule argument

Additional information

Used terms: h2

Proof step 39 from proof introduce-sharing (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 40:

  h ⊥ h2, h0 ⊥ h2, h0 ⊥ h, hP(h2), hP0(h2 ∪ h0), hP1(h2 ∪ h), ∀ h1, h3, h4. ¬ h1 ⊥ h4 ∨ ¬ h1 ⊥ h3 ∨ ¬ h3 ⊥ h4 ∨ h0 ∪ h ∪ h2 ≠ h1 ∪ h3 ∪ h4 ∨ ¬ hP0(h1 ∪ h4) ∨ ¬ hP1(h3 ∪ h4) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss)

Proof step 40 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 41:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 42:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, hP(h2), hP0(h2 ∪ h0), hP1(h ∪ h2), ∀ h1, h3, h4. ¬ h1 ⊥ h4 ∨ ¬ h1 ⊥ h3 ∨ ¬ h3 ⊥ h4 ∨ h0 ∪ h ∪ h2 ≠ h1 ∪ h3 ∪ h4 ∨ ¬ hP0(h1 ∪ h4) ∨ ¬ hP1(h3 ∪ h4) ⊦

Rule argument

Additional information

Used terms: h, h2

Proof step 41 from proof introduce-sharing (previous step) (back to tree)

A spec-lemma

Proof step 42 from proof introduce-sharing (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 43:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 44:

  h ⊥ h2, h0 ⊥ h, h0 ⊥ h2, hP(h2), hP0(h0 ∪ h2), hP1(h ∪ h2), ∀ h1, h3, h4. ¬ h1 ⊥ h4 ∨ ¬ h1 ⊥ h3 ∨ ¬ h3 ⊥ h4 ∨ h0 ∪ h ∪ h2 ≠ h1 ∪ h3 ∪ h4 ∨ ¬ hP0(h1 ∪ h4) ∨ ¬ hP1(h3 ∪ h4) ⊦

Rule argument

Additional information

Used terms: h0, h2

Proof step 43 from proof introduce-sharing (previous step) (back to tree)

A spec-lemma

Proof step 44 from proof introduce-sharing (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: h0, h, h2