Proof step 1 from proof ex1-shift (conclusion) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 2:

  ((_∃ rP * hP))(h) ⊦ ∃ r. ((rP(r) * hP))(h)

• Proof step 6:

  ¬ ((_∃ rP * hP))(h), ∃ r. ((rP(r) * hP))(h) ⊦

Rule argument

Additional information

Proof step 2 from proof ex1-shift (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. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))

• Proof step 4:

  ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ (_∃ rP)(h0) ∧ hP(h1) ⊦ ∃ r. ((rP(r) * hP))(h)

Rule argument

Additional information

Used terms: h, hP, _∃ rP

Proof step 3 from proof ex1-shift (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof ex1-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  hP(h1), rP(r)(h0), h0 ⊥ h1 ⊦ ∃ r. ((rP(r) * hP))(h0 ∪ h1)

Rule argument

Additional information

Used simplifier rules: ex1: ⊦ (_∃ rP)(h) ↔ (∃ r. rP(r)(h));

Proof step 5 from proof ex1-shift (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → ((hP0 * hP1))(h0 ∪ h1); (ss) sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep Used terms: r

Proof step 6 from proof ex1-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  ((rP(r) * hP))(h), ¬ ((_∃ rP * hP))(h) ⊦

Rule argument

Additional information

Proof step 7 from proof ex1-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

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

• Proof step 9:

  ¬ ((_∃ rP * hP))(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ rP(r)(h0) ∧ hP(h1) ⊦

Rule argument

Additional information

Used terms: h, hP, rP(r)

Proof step 8 from proof ex1-shift (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof ex1-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  hP(h1), rP(r)(h0), h0 ⊥ h1, ¬ ((_∃ rP * hP))(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 10 from proof ex1-shift (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

  hP(h1), rP(r)(h0), h0 ⊥ h1, ¬ ((_∃ rP * hP))(h0 ∪ h1) ⊦ (_∃ rP)(h0) ∧ hP(h1) ∧ h0 ⊥ h1

Rule argument

Additional information

Used terms: h1, h0, hP, _∃ rP

Proof step 11 from proof ex1-shift (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof ex1-shift (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  hP(h1), rP(r)(h0), h0 ⊥ h1, ¬ ((_∃ rP * hP))(h0 ∪ h1) ⊦ ∃ r. rP(r)(h0)

Rule argument

Additional information

Used simplifier rules: ex1: ⊦ (_∃ rP)(h) ↔ (∃ r. rP(r)(h));

Proof step 13 from proof ex1-shift (previous step) (back to tree)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

Additional information

Used terms: r