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:

  ((hP1 -* hP2))(h1), hP0(h0), h0 ⊥ 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:

  ((hP1 -* hP2))(h1), hP0(h0), h0 ⊥ 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:

  ((hP1 -* hP2))(h1), hP0(h0), hP1(h2), h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ 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:

  hP0(h0), hP1(h2), h2 ⊥ h1, h2 ⊥ h0, h0 ⊥ h1, ¬ ((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:

  hP0(h0), hP1(h2), h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ h1, ¬ ((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:

  hP0(h0), hP1(h2), hP2(h2 ∪ h1), h2 ⊥ h1, h2 ⊥ h0, h0 ⊥ 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:

  hP0(h0), hP1(h2), hP2(h2 ∪ h1), h0 ⊥ h1, h2 ⊥ h0, 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 obtained the following new goals

• Proof step 16:

  hP0(h0), hP1(h2), hP2(h2 ∪ h1), h2 ⊥ h1, h2 ⊥ h0, h0 ⊥ h1 ⊦ h0 ⊥ h2 ∪ h1 ∧ h2 ∪ h0 ∪ h1 = h0 ∪ h2 ∪ h1 ∧ hP0(h0) ∧ hP2(h2 ∪ h1)

Rule argument

Additional information

Used terms: h0, h2 ∪ h1

Proof step 16 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  h2 ∪ h0 ∪ h1 ≠ h0 ∪ h2 ∪ h1, hP0(h0), hP1(h2), hP2(h2 ∪ h1), h0 ⊥ h1, h2 ⊥ h0, h2 ⊥ 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 17 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

• Proof step 19:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h1 ⊥ h2 → (h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2); (s) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h0, h2 ∪ h1

Proof step 18 from proof wand-shift (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof wand-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h1 ⊥ h2 → (h1 ∪ h2 ⊥ h0 ↔ h0 ⊥ h1 ∧ h0 ⊥ h2); (s)

Proof step 20 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

• Proof step 22:

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

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h2, h1, h0

Proof step 21 from proof wand-shift (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof wand-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 23:

  ⊦ h0 ⊥ h1 → h1 ∪ h0 = h0 ∪ h1

Rule argument

Additional information

Used terms: h1, h0

Proof step 23 from proof wand-shift (previous step) (back to tree)

A spec-lemma