Proof step 1 from proof propagate (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬((!(id, hO2, hP1) * true)(h2) ↔ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2)), id ≠ me, (!(me, hO2, hP) * hP0)(h1), rely(me, hO1, h1, hO2, h2) ⊦

Rule argument

Additional information

Proof step 2 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  id ≠ me, (!(me, hO2, hP) * hP0)(h1), (!(id, hO2, hP1) * true)(h2), rely(me, hO1, h1, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2) ⊦

Rule argument

Additional information

Used simplifier rules: sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep ⊦ (hP0 * hP)(h) → (true * hP)(h); (s)

Proof step 3 from proof propagate (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:

  id ≠ me, (!(id, hO2, hP1) * true)(h2), rely(me, hO1, h1, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ∃ h0, h2. h0 ⊥ h2 ∧ h1 = h0 ∪ h2 ∧ !(me, hO2, hP)(h0) ∧ hP0(h2) ⊦

Rule argument

Additional information

Used terms: h1, hP0, !(me, hO2, hP)

Proof step 4 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 5 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 6:

  id ≠ me, !(me, hO2, hP)(h0), (!(id, hO2, hP1) * true)(h2), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2) ⊦

Rule argument

Additional information

Proof step 6 from proof propagate (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 7:

  id ≠ me, (!(id, hO2, hP1) * true)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h0 ⊆ h2, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2) ⊦

• Proof step 43:

  id ≠ me, (!(id, hO2, hP1) * true)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ¬ h0 ⊆ h2 ⊦

Rule argument

Additional information

Used formulas: h0 ⊆ h2

Proof step 7 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

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

• Proof step 9:

  id ≠ me, !(me, hO2, hP)(h0), (!(id, hO2, hP1) * true)(h2), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ∃ h. h0 ∪ h = h2 ∧ h ⊥ h0 ⊦

Rule argument

Additional information

Used terms: h2, h0

Proof step 8 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  id ≠ me, (!(id, hO2, hP1) * true)(h0 ∪ h1), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h0 ∪ h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 10 from proof propagate (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:

  id ≠ me, !(me, hO2, hP)(h0), (!(id, hO2, hP1) * true)(h0 ∪ h1), h0 ⊥ h, h1 ⊥ h0, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h0 ∪ h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h0 ∪ h1) ⊦ !(me, hO2, hP)(h0) ∧ (!(id, hO2, hP1) * true)(h1) ∧ h0 ⊥ h1

Rule argument

Additional information

Used terms: h1, h0, !(id, hO2, hP1) * true, !(me, hO2, hP)

Proof step 11 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  id ≠ me, (!(id, hO2, hP1) * true)(h0 ∪ h1), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h0 ∪ h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h0 ∪ h1), ¬ (!(id, hO2, hP1) * true)(h1) ⊦

Rule argument

Additional information

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

Proof step 13 from proof propagate (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:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h0 ∪ h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h0 ∪ h1), ¬ (!(id, hO2, hP1) * true)(h1), ∃ h2, h3. h2 ⊥ h3 ∧ h0 ∪ h1 = h2 ∪ h3 ∧ !(id, hO2, hP1)(h2) ∧ true(h3) ⊦

Rule argument

Additional information

Used terms: h0 ∪ h1, true, !(id, hO2, hP1)

Proof step 14 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ true(h); (s)

Proof step 16 from proof propagate (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 17:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, h2 ⊆ h1, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1) ⊦

• Proof step 22:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ h2 ⊆ h1 ⊦

Rule argument

Additional information

Used formulas: h2 ⊆ h1

Proof step 17 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 18:

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

• Proof step 19:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ∃ h. h2 ∪ h = h1 ∧ h ⊥ h2 ⊦

Rule argument

Additional information

Used terms: h1, h2

Proof step 18 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 19 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 20:

  h0 ∪ h2 ∪ h4 = h2 ∪ h3, id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h0 ⊥ h2, h0 ⊥ h4, h2 ⊥ h3, h4 ⊥ h2, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h2 ∪ h4) ⊦

Rule argument

Additional information

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

Proof step 20 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 21:

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

Rule argument

Additional information

Used simplifier rules: ⊦ true(h); (s) ⊦ h0 ⊥ h1 → h1 ⊥ h0; (ss) Used terms: h4, h2, true, !(id, hO2, hP1)

Proof step 21 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 22 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 23:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 24:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ∃ r. ¬((r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))) ⊦

Rule argument

Additional information

Used terms: h1, h2

Proof step 23 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 24 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  h0 ∪ h1 = h2 ∪ h3, id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Proof step 25 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

  ⊦ h0 = h1 ↔ (∀ r. (r ∈ h0 ↔ r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 27:

  id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ∀ r. (r ∈ h0 ∪ h1 ↔ r ∈ h2 ∪ h3) ∧ (r ∈ h0 ∪ h1 → (h0 ∪ h1)(r) = (h2 ∪ h3)(r)) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used terms: h2 ∪ h3, h0 ∪ h1

Proof step 26 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ∀ r0. (r0 ∈ h0 ∪ h1 ↔ r0 ∈ h2 ∨ r0 ∈ h3) ∧ (r0 ∈ h0 ∪ h1 → (h0 ∪ h1)(r0) = (h2 ∪ h3)(r0)) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

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

Proof step 28 from proof propagate (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 29:

  r ∈ h0 ∪ h1 ↔ r ∈ h2 ∨ r ∈ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h0 ∪ h1 → (h0 ∪ h1)(r) = (h2 ∪ h3)(r) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used terms: r

Proof step 29 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  r ∈ h0 ∪ h1 ↔ r ∈ h2 ∨ r ∈ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), r ∈ h2 ∨ r ∈ h3 → (h0 ∪ h1)(r) = (r ∈ h2 ⊃ h2(r);h3(r)) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used simplifier rules: r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

Proof step 30 from proof propagate (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 31:

  (h0 ∪ h1)(r) = (r ∈ h2 ⊃ h2(r);h3(r)), r ∈ h0 ∪ h1 ↔ r ∈ h2 ∨ r ∈ h3, id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Proof step 31 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 32:

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

• Proof step 33:

  (h0 ∪ h1)(r) = (r ∈ h2 ⊃ h2(r);h3(r)), r ∈ h0 ∪ h1 ↔ r ∈ h2 ∨ r ∈ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r)), r ∈ h0 ∪ h1 ∧ h0 ⊥ h1

• Proof step 36:

  (r ∈ h0 ⊃ h0(r);h1(r)) = (r ∈ h2 ⊃ h2(r);h3(r)), r ∈ h0 ∪ h1 ↔ r ∈ h2 ∨ r ∈ h3, id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h0 ∪ h1, rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used terms: r, h1, h0

Proof step 32 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 33 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  (h0 ∪ h1)(r) = h2(r), id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h2, r ∈ h0 ∪ h1, rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ r ∈ h0, ¬ r ∈ h1 ⊦

Rule argument

Additional information

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

Proof step 34 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1)

Rule argument

Additional information

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

Proof step 35 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  (r ∈ h0 ⊃ h0(r);h1(r)) = (r ∈ h2 ⊃ h2(r);h3(r)), id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h0, h2 ⊥ h3, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h2 ∨ r ∈ h3, r ∈ h0 ∨ r ∈ h1 ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

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

Proof step 37 from proof propagate (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 38:

  h0(r) = (r ∈ h2 ⊃ h2(r);h3(r)), id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h1, h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h0, rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h0 ∨ r ∈ h1, r ∈ h2 ∨ r ∈ h3 ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Proof step 38 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 39:

  h0(r) = (r ∈ h2 ⊃ h2(r);h3(r)), id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h0, rely(me, hO1, h0 ∪ h, hO2, h2 ∪ h3), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h2 ∨ r ∈ h3 ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Proof step 39 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

  ⊦ rely(me, of1, h1, of2, h2) ↔ (∀ r. (of1(r) = me ↔ of2(r) = me) ∧ (of1(r) = me → (r ∈ h1 ↔ r ∈ h2) ∧ (r ∈ h1 → h1(r) = h2(r))))

• Proof step 41:

  h0(r) = (r ∈ h2 ⊃ h2(r);h3(r)), id ≠ me, !(me, hO2, hP)(h0), !(id, hO2, hP1)(h2), h0 ⊥ h1, h0 ⊥ h, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h0, ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h2 ∨ r ∈ h3, ∀ r. (hO1(r) = me ↔ hO2(r) = me) ∧ (hO1(r) = me → (r ∈ h0 ∪ h ↔ r ∈ h2 ∪ h3) ∧ (r ∈ h0 ∪ h → (h0 ∪ h)(r) = (h2 ∪ h3)(r))) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used terms: h2 ∪ h3, hO1, h0 ∪ h, me, hO2

Proof step 40 from proof propagate (previous step) (back to tree)

Lemma rely

Proof step 41 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  h0(r) = (r ∈ h2 ⊃ h2(r);h3(r)), id ≠ me, !(id, hO2, hP1)(h2), !(me, hO2, hP)(h0), h0 ⊥ h, h0 ⊥ h1, h1 ⊥ h0, h2 ⊥ h3, hP0(h), r ∈ h0, ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2 ∪ h3), ¬ (!(id, hO2, hP1) * true)(h1), r ∈ h2 ∨ r ∈ h3, ∀ r0. (hO1(r0) = me ↔ hO2(r0) = me) ∧ (hO1(r0) = me → (r0 ∈ h0 ∨ r0 ∈ h ↔ r0 ∈ h2 ∨ r0 ∈ h3) ∧ (r0 ∈ h0 ∨ r0 ∈ h → (r0 ∈ h0 ⊃ h0(r0);h(r0)) = (h2 ∪ h3)(r0))) ⊦ (r ∈ h2 → r ∈ h1) ∧ (r ∈ h2 → h2(r) = h1(r))

Rule argument

Additional information

Used simplifier rules: r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

Proof step 42 from proof propagate (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: owned-r: ⊦ !(me, hO, hP)(h) ∧ hO(r) ≠ me → ¬ r ∈ h; ⊦ 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) Used terms: r

Proof step 43 from proof propagate (previous step) (back to tree)

User applied rule weakening

And obtained the following new goals

• Proof step 44:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ¬ h0 ⊆ h2 ⊦

Rule argument

Additional information

Proof step 44 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 45:

  ⊦ h0 ⊆ h1 ↔ (∀ r. (r ∈ h0 → r ∈ h1) ∧ (r ∈ h0 → h0(r) = h1(r)))

• Proof step 46:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ∃ r. ¬((r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))) ⊦

Rule argument

Additional information

Used terms: h2, h0

Proof step 45 from proof propagate (previous step) (back to tree)

A spec-lemma

Proof step 46 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), rely(me, hO1, h0 ∪ h, hO2, h2), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2) ⊦ (r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))

Rule argument

Additional information

Proof step 47 from proof propagate (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 48:

  ⊦ rely(me, of1, h1, of2, h2) ↔ (∀ r. (of1(r) = me ↔ of2(r) = me) ∧ (of1(r) = me → (r ∈ h1 ↔ r ∈ h2) ∧ (r ∈ h1 → h1(r) = h2(r))))

• Proof step 49:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ∀ r. (hO1(r) = me ↔ hO2(r) = me) ∧ (hO1(r) = me → (r ∈ h0 ∪ h ↔ r ∈ h2) ∧ (r ∈ h0 ∪ h → (h0 ∪ h)(r) = h2(r))) ⊦ (r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))

Rule argument

Additional information

Used terms: h2, hO1, h0 ∪ h, me, hO2

Proof step 48 from proof propagate (previous step) (back to tree)

Lemma rely

Proof step 49 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 50:

  id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), ∀ r0. (hO1(r0) = me ↔ hO2(r0) = me) ∧ (hO1(r0) = me → (r0 ∈ h0 ∨ r0 ∈ h ↔ r0 ∈ h2) ∧ (r0 ∈ h0 ∨ r0 ∈ h → (r0 ∈ h0 ⊃ h0(r0);h(r0)) = h2(r0))) ⊦ (r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))

Rule argument

Additional information

Used simplifier rules: r ∈ h0 ∪ h1 ⊦ h0 ⊥ h1 → (h0 ∪ h1)(r) = (r ∈ h0 ⊃ h0(r);h1(r)); (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s)

Proof step 50 from proof propagate (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 51:

  hO1(r) = me ↔ hO2(r) = me, id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), hO1(r) = me → (r ∈ h0 ∨ r ∈ h ↔ r ∈ h2) ∧ (r ∈ h0 ∨ r ∈ h → (r ∈ h0 ⊃ h0(r);h(r)) = h2(r)), ∀ r0. (hO1(r0) = me ↔ hO2(r0) = me) ∧ (hO1(r0) = me → (r0 ∈ h0 ∨ r0 ∈ h ↔ r0 ∈ h2) ∧ (r0 ∈ h0 ∨ r0 ∈ h → (r0 ∈ h0 ⊃ h0(r0);h(r0)) = h2(r0))) ⊦ (r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))

Rule argument

Additional information

Used terms: r

Proof step 51 from proof propagate (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 52:

  hO1(r) = me ↔ hO2(r) = me, id ≠ me, !(me, hO2, hP)(h0), h0 ⊥ h, hP0(h), ¬ (!(me, hO2, hP) * !(id, hO2, hP1) * true)(h2), hO2(r) = me → (r ∈ h0 ∨ r ∈ h ↔ r ∈ h2) ∧ (r ∈ h0 ∨ r ∈ h → (r ∈ h0 ⊃ h0(r);h(r)) = h2(r)), ∀ r0. (hO1(r0) = me ↔ hO2(r0) = me) ∧ (hO1(r0) = me → (r0 ∈ h0 ∨ r0 ∈ h ↔ r0 ∈ h2) ∧ (r0 ∈ h0 ∨ r0 ∈ h → (r0 ∈ h0 ⊃ h0(r0);h(r0)) = h2(r0))) ⊦ (r ∈ h0 → r ∈ h2) ∧ (r ∈ h0 → h0(r) = h2(r))

Rule argument

Additional information

Proof step 52 from proof propagate (previous step) (back to tree)

User applied rule case distinction

And closed the goal!

Rule argument

Additional information

Used simplifier rules: owned-r: ⊦ !(me, hO, hP)(h) ∧ hO(r) ≠ me → ¬ r ∈ h; owned-r: ⊦ !(me, hO, hP)(h) ∧ hO(r) ≠ me → ¬ r ∈ h;