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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(h(r) = o1 ↔ o0 = o1), (r => o0 ⊗ hP)(h) ⊦

Rule argument

Additional information

Proof step 2 from proof at-eq-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:

  ¬(h(r) = o1 ↔ o0 = o1), ∃ h0, h1, h2. h0 ⊥ h2 ∧ h0 ⊥ h1 ∧ h1 ⊥ h2 ∧ h = h0 ∪ h1 ∪ h2 ∧ (r => o0)(h0 ∪ h2) ∧ hP(h1 ∪ h2) ⊦

Rule argument

Additional information

Used simplifier rules: sharing-ass: ⊦ hP0 ⊗ hP1 ⊗ hP2 = (hP0 ⊗ hP1) ⊗ hP2; from specification map-sep sharing-com: ⊦ hP0 ⊗ hP1 = hP1 ⊗ hP0; from specification map-sep Used terms: h, r => o0, hP

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  h0 ∪ h2 = ∅(r := o0), ¬((h0 ∪ h1 ∪ h2)(r) = o1 ↔ o0 = o1), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP(h1 ∪ h2) ⊦

Rule argument

Additional information

Used simplifier rules: maplet: ⊦ (r => o)(h) ↔ h = ∅(r := o);

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

  r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r))

• Proof step 7:

  h0 ∪ h2 = ∅(r := o0), ¬((h0 ∪ h1 ∪ h2)(r) = o1 ↔ o0 = o1), h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP(h1 ∪ h2) ⊦ r ∈ h0 ∪ h1 ∪ h2 ∧ h0 ⊥ h1 ∪ h2

• Proof step 12:

  h0 ∪ h2 = ∅(r := o0), ¬((r ∈ h0 ⊃ h0(r);(h1 ∪ h2)(r)) = o1 ↔ o0 = o1), h0 ⊥ h1, h0 ⊥ h2, h0 ⊥ h1 ∪ h2, h1 ⊥ h2, hP(h1 ∪ h2), r ∈ h0 ∪ h1 ∪ h2 ⊦

Rule argument

Additional information

Used terms: h0, h1 ∪ h2, r

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  h0 ∪ h2 = ∅(r := o0), ¬((h0 ∪ h1 ∪ h2)(r) = o1 ↔ o0 = o1), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP(h1 ∪ h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2 ⊦

Rule argument

Additional information

Used simplifier rules: h1 ⊥ h ⊦ h0 ⊥ h1 ∪ h ↔ h0 ⊥ h1 ∧ h0 ⊥ h; (ws) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 9:

  ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1(a) = st2(a))

• Proof step 10:

  ¬((h0 ∪ h1 ∪ h2)(r) = o1 ↔ o0 = o1), h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP(h1 ∪ h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ∀ a. (a ∈ h0 ∪ h2 ↔ a ∈ ∅(r := o0)) ∧ (h0 ∪ h2)(a) = ∅(r := o0)(a) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h2, ∅(r := o0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  ¬((h0 ∪ h1 ∪ h2)(r) = o1 ↔ o0 = o1), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP(h1 ∪ h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ∀ a. (a ∈ h0 ∨ a ∈ h2 ↔ a = r) ∧ (h0 ∪ h2)(a) = ∅(r := o0)(a) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ ∅(b := d) ↔ a = b; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used terms: r

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h0 ∪ h2 = ∅(r := o0), ¬((r ∈ h0 ⊃ h0(r);(h1 ∪ h2)(r)) = o1 ↔ o0 = o1), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP(h1 ∪ h2), r ∈ h0 ∪ h1 ∪ h2 ⊦

Rule argument

Additional information

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

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1(a) = st2(a))

• Proof step 15:

  ¬((r ∈ h0 ⊃ h0(r);(h1 ∪ h2)(r)) = o1 ↔ o0 = o1), h0 ⊥ h1, h0 ⊥ h2, h1 ⊥ h2, hP(h1 ∪ h2), r ∈ h0 ∪ h1 ∪ h2, ∀ a. (a ∈ h0 ∪ h2 ↔ a ∈ ∅(r := o0)) ∧ (h0 ∪ h2)(a) = ∅(r := o0)(a) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h2, ∅(r := o0)

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  ¬((r ∈ h0 ⊃ h0(r);(h1 ∪ h2)(r)) = o1 ↔ o0 = o1), h0 ⊥ h2, h0 ⊥ h1, h1 ⊥ h2, hP(h1 ∪ h2), r ∈ h0 ∪ h1 ∪ h2, ∀ a. (a ∈ h0 ∨ a ∈ h2 ↔ a = r) ∧ (h0 ∪ h2)(a) = ∅(r := o0)(a) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ ∅(b := d) ↔ a = b; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

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

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ a = b → st(b := d)(a) = d; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s) ⊦ r ∈ h1 ∧ h0 ⊥ h1 → (r ∈ h0 ↔ false); (s) Used terms: r