Proof step 1 from proof ex-shift-o-p (conclusion) (back to tree)

⊦ (!(me, hO, _∃ ohP * hP0) * hP)(h) ↔ (∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h))

User applied rule case distinction

And obtained the following new goals

Proof step 2:
(!(me, hO, _∃ ohP * hP0) * hP)(h) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h)
Proof step 19:
¬ (!(me, hO, _∃ ohP * hP0) * hP)(h), ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h) ⊦

Rule argument

pos: right 1

Additional information

Proof step 2 from proof ex-shift-o-p (previous step) (back to tree)

(!(me, hO, _∃ ohP * hP0) * hP)(h) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h)

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 ∧ !(me, hO, _∃ ohP * hP0)(h0) ∧ hP(h1) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h)

Rule argument

spec: heap-sep, name: sep, seq: ⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1)), subst:[hP0, hP1, h] → [!(me, hO, _∃ ohP * hP0), hP, h]; [A] → [A], <all paths>

Additional information

Used terms: !(me, hO, _∃ ohP * hP0), hP, h

Proof step 3 from proof ex-shift-o-p (previous step) (back to tree)

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

A spec-lemma

Proof step 4 from proof ex-shift-o-p (previous step) (back to tree)

∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ !(me, hO, _∃ ohP * hP0)(h0) ∧ hP(h1) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
!(me, hO, _∃ ohP * hP0)(h0), h0 ⊥ h1, hP(h1) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

Rule argument

Additional information

Proof step 5 from proof ex-shift-o-p (previous step) (back to tree)

!(me, hO, _∃ ohP * hP0)(h0), h0 ⊥ h1, hP(h1) ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)
Proof step 7:
(_∃ ohP * hP0)(h0), h0 ⊥ h1, hP(h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

Rule argument

spec: <toplevel>, name: owned, seq: ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id), subst:[id, of, hP, h] → [me, hO, _∃ ohP * hP0, h0]; [A] → [A], <all paths>

Additional information

Used terms: me, hO, _∃ ohP * hP0, h0

Proof step 6 from proof ex-shift-o-p (previous step) (back to tree)

⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

Lemma owned

Proof step 7 from proof ex-shift-o-p (previous step) (back to tree)

(_∃ ohP * hP0)(h0), h0 ⊥ h1, hP(h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

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 ⊦ (_∃ bP * hP)(h) ↔ (∃ b. (bP(b) * hP)(h)); (s)

Proof step 8 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))
Proof step 10:
h0 ⊥ h1, hP(h1), ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ ohP(o)(h1) ∧ hP0(h2), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

Rule argument

spec: heap-sep, name: sep, seq: ⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1)), subst:[h, hP1, hP0] → [h0, hP0, ohP(o)], <all paths>

Additional information

Used terms: h0, hP0, ohP(o)

Proof step 9 from proof ex-shift-o-p (previous step) (back to tree)

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

A spec-lemma

Proof step 10 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ ohP(o)(h1) ∧ hP0(h2), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h0 ∪ h1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

Additional information

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

Proof step 11 from proof ex-shift-o-p (previous step) (back to tree)

h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

User applied rule exists right

And obtained the following new goals

Proof step 12:
h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

pos: right 1 terms: o abort: F discard: F computed: T

Additional information

Used terms: o

Proof step 12 from proof ex-shift-o-p (previous step) (back to tree)

h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)
Proof step 14:
h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ !(me, hO, ohP(o) * hP0)(h ∪ h2) ∧ hP(h1) ∧ h ∪ h2 ⊥ h1, ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

spec: heap-sep, name: split-union, seq: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1), subst:[h1, h0, hP1, hP0] → [h1, h ∪ h2, hP, !(me, hO, ohP(o) * hP0)], <all paths>

Additional information

Used terms: h1, h ∪ h2, hP, !(me, hO, ohP(o) * hP0)

Proof step 13 from proof ex-shift-o-p (previous step) (back to tree)

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

A spec-lemma

Proof step 14 from proof ex-shift-o-p (previous step) (back to tree)

h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ !(me, hO, ohP(o) * hP0)(h ∪ h2) ∧ hP(h1) ∧ h ∪ h2 ⊥ h1, ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ¬ !(me, hO, ohP(o) * hP0)(h ∪ h2), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

Additional information

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

Proof step 15 from proof ex-shift-o-p (previous step) (back to tree)

h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ¬ !(me, hO, ohP(o) * hP0)(h ∪ h2), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)
Proof step 17:
h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ¬ (ohP(o) * hP0)(h ∪ h2) ∨ (∃ r. ¬(r ∈ h ∪ h2 → hO(r) = me)), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

spec: <toplevel>, name: owned, seq: ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id), subst:[h, hP, of, id] → [h ∪ h2, ohP(o) * hP0, hO, me], <all paths>

Additional information

Used terms: h ∪ h2, ohP(o) * hP0, hO, me

Proof step 16 from proof ex-shift-o-p (previous step) (back to tree)

⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

Lemma owned

Proof step 17 from proof ex-shift-o-p (previous step) (back to tree)

h ⊥ h2, h1 ⊥ h, h1 ⊥ h2, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ¬ (ohP(o) * hP0)(h ∪ h2) ∨ (∃ r. ¬(r ∈ h ∪ h2 → hO(r) = me)), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
hO(r) ≠ me, h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me, r ∈ h ∨ r ∈ h2 ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

Rule argument

Additional information

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

Proof step 18 from proof ex-shift-o-p (previous step) (back to tree)

hO(r) ≠ me, h ⊥ h2, h1 ⊥ h2, h1 ⊥ h, hP(h1), hP0(h2), ohP(o)(h), ¬ (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1), ∀ r. r ∈ h ∨ r ∈ h2 → hO(r) = me, r ∈ h ∨ r ∈ h2 ⊦ ∃ o. (!(me, hO, ohP(o) * hP0) * hP)((h ∪ h2) ∪ h1)

User applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: r abort: F discard: F computed: T

Additional information

Used terms: r

Proof step 19 from proof ex-shift-o-p (previous step) (back to tree)

¬ (!(me, hO, _∃ ohP * hP0) * hP)(h), ∃ o. (!(me, hO, ohP(o) * hP0) * hP)(h) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
(!(me, hO, ohP(o) * hP0) * hP)(h), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h) ⊦

Rule argument

Additional information

Proof step 20 from proof ex-shift-o-p (previous step) (back to tree)

(!(me, hO, ohP(o) * hP0) * hP)(h), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 21:
⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1))
Proof step 22:
¬ (!(me, hO, _∃ ohP * hP0) * hP)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ !(me, hO, ohP(o) * hP0)(h0) ∧ hP(h1) ⊦

Rule argument

spec: heap-sep, name: sep, seq: ⊦ (hP0 * hP1)(h) ↔ (∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ hP0(h0) ∧ hP1(h1)), subst:[h, hP1, hP0] → [h, hP, !(me, hO, ohP(o) * hP0)], <all paths>

Additional information

Used terms: h, hP, !(me, hO, ohP(o) * hP0)

Proof step 21 from proof ex-shift-o-p (previous step) (back to tree)

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

A spec-lemma

Proof step 22 from proof ex-shift-o-p (previous step) (back to tree)

¬ (!(me, hO, _∃ ohP * hP0) * hP)(h), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ !(me, hO, ohP(o) * hP0)(h0) ∧ hP(h1) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1) ⊦

Rule argument

Additional information

Proof step 23 from proof ex-shift-o-p (previous step) (back to tree)

!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 24:
⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1)
Proof step 25:
!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1) ⊦ !(me, hO, _∃ ohP * hP0)(h0) ∧ hP(h1) ∧ h0 ⊥ h1

Rule argument

spec: heap-sep, name: split-union, seq: ⊦ hP0(h0) ∧ hP1(h1) ∧ h0 ⊥ h1 → (hP0 * hP1)(h0 ∪ h1), subst:[hP0, hP1, h0, h1] → [!(me, hO, _∃ ohP * hP0), hP, h0, h1]; [A] → [A], <all paths>

Additional information

Used terms: !(me, hO, _∃ ohP * hP0), hP, h0, h1

Proof step 24 from proof ex-shift-o-p (previous step) (back to tree)

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

A spec-lemma

Proof step 25 from proof ex-shift-o-p (previous step) (back to tree)

!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1) ⊦ !(me, hO, _∃ ohP * hP0)(h0) ∧ hP(h1) ∧ h0 ⊥ h1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0) ⊦

Rule argument

Additional information

Proof step 26 from proof ex-shift-o-p (previous step) (back to tree)

!(me, hO, ohP(o) * hP0)(h0), h0 ⊥ h1, hP(h1), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0) ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)
Proof step 28:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

spec: <toplevel>, name: owned, seq: ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id), subst:[h, hP, of, id] → [h0, ohP(o) * hP0, hO, me], <all paths>

Additional information

Used terms: h0, ohP(o) * hP0, hO, me

Proof step 27 from proof ex-shift-o-p (previous step) (back to tree)

⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

Lemma owned

Proof step 28 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Proof step 29 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ !(me, hO, _∃ ohP * hP0)(h0), ∀ r. r ∈ h0 → hO(r) = me ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)
Proof step 31:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ (_∃ ohP * hP0)(h0) ∨ (∃ r. ¬(r ∈ h0 → hO(r) = me)), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

spec: <toplevel>, name: owned, seq: ⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id), subst:[id, of, hP, h] → [me, hO, _∃ ohP * hP0, h0]; [A] → [A], <all paths>

Additional information

Used terms: me, hO, _∃ ohP * hP0, h0

Proof step 30 from proof ex-shift-o-p (previous step) (back to tree)

⊦ !(id, of, hP)(h) ↔ hP(h) ∧ (∀ r. r ∈ h → of(r) = id)

Lemma owned

Proof step 31 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬ (_∃ ohP * hP0)(h0) ∨ (∃ r. ¬(r ∈ h0 → hO(r) = me)), ∀ r. r ∈ h0 → hO(r) = me ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬(∃ b. (ohP(b) * hP0)(h0)) ∨ ¬(r ∈ h0 → hO(r) = me), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

Additional information

Proof step 32 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ¬(∃ b. (ohP(b) * hP0)(h0)) ∨ ¬(r ∈ h0 → hO(r) = me), ∀ r. r ∈ h0 → hO(r) = me ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 33:
h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ b. (ohP(b) * hP0)(h0)
Proof step 34:
hO(r) ≠ me, h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), r ∈ h0, ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Rule argument

pos: left 5

Additional information

Proof step 33 from proof ex-shift-o-p (previous step) (back to tree)

h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦ ∃ b. (ohP(b) * hP0)(h0)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: o

Additional information

Used terms: o

Proof step 34 from proof ex-shift-o-p (previous step) (back to tree)

hO(r) ≠ me, h0 ⊥ h1, hP(h1), (ohP(o) * hP0)(h0), r ∈ h0, ¬ (!(me, hO, _∃ ohP * hP0) * hP)(h0 ∪ h1), ∀ r. r ∈ h0 → hO(r) = me ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: r

Additional information

Used terms: r

Proof Tree for ex-shift-o-p (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.