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

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  hP(∅), ¬ hP(h), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r) ⊦

Rule argument

Additional information

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

User applied rule induction

And obtained the following new goals

• Proof step 3:

  hP(∅), ¬ hP(h), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h0, hP. # h0 < # h → hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used simplifier rules: <; (wellfounded) Used terms: # h

Proof step 3 from proof structind (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 4:

  h ≠ ∅, hP(∅), ¬ hP(h), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h0, hP. # h0 < # h → hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used formulas: h = ∅

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

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

• Proof step 6:

  hP(∅), ¬ hP(h), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∃ r. ¬((r ∈ h ↔ r ∈ ∅) ∧ (r ∈ h → h[r] = ∅[r])), ∀ h0, hP. # h0 < # h → hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used terms: h, ∅

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

Lemma ext

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  hP(∅), r ∈ h, ¬ hP(h), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h0, hP. # h0 < # h → hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used simplifier rules: in-emp: ⊦ ¬ r ∈ ∅;

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

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ r ≠ null → (r ∈ h ↔ (∃ h0, o. h = (h0 ++ r)[r, o] ∧ ¬ r ∈ h0))

• Proof step 9:

  r ≠ null, hP(∅), ¬ hP(h), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∃ h0, o. h = (h0 ++ r)[r, o] ∧ ¬ r ∈ h0, ∀ h0, hP. # h0 < # h → hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used simplifier rules: sanity: ⊦ ¬ null ∈ h; Used terms: r, h

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

Lemma exin

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  r ≠ null, hP(∅), ¬ hP((h0 ++ r)[r, o]), ¬ r ∈ h0, ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h1, hP. # h1 < # (h0 ++ r)[r, o] → hP(∅) ∧ ¬ hP(h1) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Proof step 10 from proof structind (previous step) (back to tree)

User applied rule apply induction

And obtained the following new goals

• Proof step 11:

  r ≠ null, hP(∅), ¬ hP((h0 ++ r)[r, o]), ¬ r ∈ h0, ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h1, hP. # h1 < # (h0 ++ r)[r, o] → hP(∅) ∧ ¬ hP(h1) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦ hP(∅) ∧ ¬ hP(h0) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r))

Rule argument

Additional information

Used simplifier rules: in-alloc: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; size-write: r ∈ h ⊦ # h[r, o] = # h; size-alloc: ¬ r ∈ h, r ≠ null ⊦ #(h ++ r) = # h + 1; ⊦ m < n + 1 ↔ ¬ n < m; (s) ⊦ ¬ n < n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic Used terms: h0, hP

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  r ≠ null, hP(∅), hP(h0), ¬ hP((h0 ++ r)[r, o]), ¬ r ∈ h0, ∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h1, hP. # h1 < # (h0 ++ r)[r, o] → hP(∅) ∧ ¬ hP(h1) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

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

User applied rule all left

And obtained the following new goals

• Proof step 13:

  r ≠ null, hP(h0), hP(∅), ¬ hP((h0 ++ r)[r, o]), ¬ r ∈ h0, hP(h0) ∧ ¬ r ∈ h0 ∧ r ≠ null → hP(h0 ++ r), ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h1, hP. # h1 < # (h0 ++ r)[r, o] → hP(∅) ∧ ¬ hP(h1) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

Used terms: h0, r

Proof step 13 from proof structind (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 14:

  r ≠ null, hP(∅), hP(h0), hP(h0 ++ r), ¬ hP((h0 ++ r)[r, o]), ¬ r ∈ h0, ∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o]), ∀ h1, hP. # h1 < # (h0 ++ r)[r, o] → hP(∅) ∧ ¬ hP(h1) ∧ (∀ h, r, o. hP(h) ∧ r ∈ h → hP(h[r, o])) ∧ (∀ h, r. hP(h) ∧ ¬ r ∈ h ∧ r ≠ null → hP(h ++ r)) → false ⊦

Rule argument

Additional information

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-alloc: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; Used terms: h0 ++ r, r, o