Proof step 1 from proof wand-shift (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (hP0 * (hP1 -* hP2))(h), ¬ (hP1 -* hP0 * hP2)(h) ⊦

Rule argument

Additional information

Proof step 2 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  (hP0 * (hP1 -* hP2))(h), ¬ (hP1 -* hP0 * hP2)(h) ⊦

Rule argument

Additional information

Proof step 3 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite 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:

  ¬ (hP1 -* hP0 * hP2)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ (hP1 -* hP2)(h1) ⊦

Rule argument

Additional information

Used terms: h, hP1 -* hP2, hP0

Proof step 4 from proof wand-shift (previous step) (back to tree)

Lemma sep

Proof step 5 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  h0 ⊥ h1, hP0(h0), (hP1 -* hP2)(h1), ¬ (hP1 -* hP0 * hP2)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 6 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

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

• Proof step 8:

  h0 ⊥ h1, hP0(h0), (hP1 -* hP2)(h1), ∃ h2. ¬(h2 ⊥ h0 ∪ h1 ∧ hP1(h2) → (hP0 * hP2)(h2 ∪ h0 ∪ h1)) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h1, hP0 * hP2, hP1

Proof step 7 from proof wand-shift (previous step) (back to tree)

Lemma wand

Proof step 8 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ h1, hP0(h0), hP1(h2), (hP1 -* hP2)(h1), ¬ (hP0 * hP2)(h2 ∪ h0 ∪ h1) ⊦

Rule argument

Additional information

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

Proof step 9 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  h0 ⊥ h1, h2 ⊥ h1, h2 ⊥ h0, hP0(h0), hP1(h2), ¬ (hP0 * hP2)(h2 ∪ h0 ∪ h1), ∀ h0. h0 ⊥ h1 ∧ hP1(h0) → hP2(h0 ∪ h1) ⊦

Rule argument

Additional information

Used terms: h1, hP2, hP1

Proof step 10 from proof wand-shift (previous step) (back to tree)

Lemma wand

Proof step 11 from proof wand-shift (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 12:

  h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ h1, hP0(h0), hP1(h2), ¬ (hP0 * hP2)(h2 ∪ h0 ∪ h1), h2 ⊥ h1 ∧ hP1(h2) → hP2(h2 ∪ h1) ⊦

Rule argument

Additional information

Used terms: h2

Proof step 12 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h0 ⊥ h1, h2 ⊥ h1, h2 ⊥ h0, hP0(h0), hP1(h2), hP2(h2 ∪ h1), ¬ (hP0 * hP2)(h2 ∪ h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 13 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

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

• Proof step 15:

  h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ h1, hP0(h0), hP1(h2), hP2(h2 ∪ h1), ∀ h3, h4. ¬(h3 ⊥ h4 ∧ h2 ∪ h0 ∪ h1 = h3 ∪ h4 ∧ hP0(h3) ∧ hP2(h4)) ⊦

Rule argument

Additional information

Used terms: h2 ∪ h0 ∪ h1, hP2, hP0

Proof step 14 from proof wand-shift (previous step) (back to tree)

Lemma sep

Proof step 15 from proof wand-shift (previous step) (back to tree)

User applied rule all left

And closed the goal!

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 ⊦ h ∪ h0 ∪ h1 = h0 ∪ h ∪ h1 ↔ true; (ss) Used terms: h0, h2 ∪ h1