Proof step 1 from proof precise-alt-def (conclusion) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 2:

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

• Proof step 3:

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

Rule argument

Additional information

Used terms: hP

Proof step 2 from proof precise-alt-def (previous step) (back to tree)

Lemma precise

Proof step 3 from proof precise-alt-def (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 4:

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

• Proof step 12:

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

Rule argument

Additional information

Proof step 4 from proof precise-alt-def (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

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

Rule argument

Additional information

Proof step 5 from proof precise-alt-def (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  ⊦ h0 ⊆ h1 ↔ (∃ h. h0 ∪ h = h1 ∧ h ⊥ h0)

• Proof step 7:

  h0 ≠ h1, h1 ⊆ h, hP(h1), hP(h0), ∀ h, h0, h1, h2. hP(h) ∧ hP(h0) ∧ h ⊥ h1 ∧ h0 ⊥ h2 ∧ h ∪ h1 = h0 ∪ h2 → h = h0, ∃ h2. h0 ∪ h2 = h ∧ h2 ⊥ h0 ⊦

Rule argument

Additional information

Used terms: h0, h

Proof step 6 from proof precise-alt-def (previous step) (back to tree)

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

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

Rule argument

Additional information

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ h0 ⊆ h1 ↔ (∃ h. h0 ∪ h = h1 ∧ h ⊥ h0)

• Proof step 10:

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

Rule argument

Additional information

Used terms: h1, h0 ∪ h2

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

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

Rule argument

Additional information

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

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

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 14:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, hP(h0), hP(h), h ⊆ h0 ∪ h2 ∧ h0 ⊆ h0 ∪ h2 ∧ hP(h) ∧ hP(h0) → h = h0 ⊦

Rule argument

Additional information

Used terms: h0 ∪ h2, h, h0

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, hP(h), hP(h0), ¬ h ⊆ h0 ∪ h2 ⊦

Rule argument

Additional information

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

Proof step 15 from proof precise-alt-def (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 16:

  ⊦ h0 ⊆ h1 ↔ (∃ h. h0 ∪ h = h1 ∧ h ⊥ h0)

• Proof step 17:

  h ∪ h1 = h0 ∪ h2, h ≠ h0, h ⊥ h1, h0 ⊥ h2, hP(h0), hP(h), ∀ h3. ¬(h ∪ h3 = h0 ∪ h2 ∧ h3 ⊥ h) ⊦

Rule argument

Additional information

Used terms: h, h0 ∪ h2

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 18:

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

Rule argument

Additional information

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

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