Proof step 1 from proof map-del (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬(((r => o * hP))(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h[r] = o) ⊦

Rule argument

Additional information

Proof step 2 from proof map-del (previous step) (back to tree)

User applied rule case distinction

And obtained the following new goals

• Proof step 3:

  ((r => o * hP))(h) ⊦ hP(h -- r) ∧ r ∈ h ∧ h[r] = o

• Proof step 6:

  h[r] = o, hP(h -- r), r ∈ h, ¬ ((r => o * hP))(h) ⊦

Rule argument

Additional information

Proof step 3 from proof map-del (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 4:

  ((r => o * hP))(h) ⊦ hP(h -- r) ∧ h[r] = o

Rule argument

Additional information

Used simplifier rules: map-in: ⊦ ((r => o * hP))(h) → r ∈ h; sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep

Proof step 4 from proof map-del (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 5:

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

Rule argument

Additional information

Used simplifier rules: maplet: ⊦ ((r => o))(h) ↔ h = (∅ ++ r)[r, o] ∧ r ≠ null; ⊦ h0 ⊥ h1 → (h0 ∪ h1) -- r = (h0 -- r) ∪ (h1 -- r); (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 ≠ null ⊦ (∅ ++ r)[r, o] ⊥ h ↔ ¬ r ∈ h; (s) r ∈ h ⊦ h[r, o] -- r = h -- r; (s) ¬ r ∈ h, r ≠ null ⊦ (h ++ r) -- r = h; (s) ⊦ ¬ r ∈ h → h -- r = h; (s) ⊦ ∅ ∪ h = h; (s) ⊦ r ∈ h[r0, o] ↔ r ∈ h; (s) ⊦ h[r0, o][r] = (r = r0 ∧ r0 ∈ h ⊃ o;h[r]); (s) ⊦ ¬ r ∈ ∅; (s) r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) Used terms: h, hP, r => o

Proof step 5 from proof map-del (previous step) (back to tree)

A spec-lemma

Proof step 6 from proof map-del (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  hP(h -- r), r ∈ h, ¬ ((r => h[r] * hP))(h) ⊦

Rule argument

Additional information

Proof step 7 from proof map-del (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:

  hP(h -- r), r ∈ h, ∀ h0, h1. ¬(h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ ((r => h[r]))(h0) ∧ hP(h1)) ⊦

Rule argument

Additional information

Used terms: h, hP, r => h[r]

Proof step 8 from proof map-del (previous step) (back to tree)

A spec-lemma

Proof step 9 from proof map-del (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  hP(h -- r), r ∈ h, ∀ h0, h1. ¬ h0 ⊥ h1 ∨ h ≠ h0 ∪ h1 ∨ h0 ≠ (∅ ++ r)[r, (h0 ∪ h1)[r]] ∨ ¬ hP(h1) ⊦

Rule argument

Additional information

Used simplifier rules: maplet: ⊦ ((r => o))(h) ↔ h = (∅ ++ r)[r, o] ∧ r ≠ null; ⊦ r ∈ h → r ≠ null; (s)

Proof step 10 from proof map-del (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: r ≠ null ⊦ (∅ ++ r)[r, o] ⊥ h ↔ ¬ r ∈ h; (s) ⊦ r ∈ h -- r0 ↔ r ≠ r0 ∧ r ∈ h; (s) ⊦ r ∈ h → r ≠ null; (s) r ≠ null ⊦ r ∈ h → (∅ ++ r)[r, h[r]] ∪ (h -- r) = h; (s) Used terms: (∅ ++ r)[r, h[r]], h -- r