Proof step 1 from proof precise-tr (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  ¬ precise(tr(r, t)) ⊦

Rule argument

Additional information

Proof step 2 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 3:

  ⊦ precise(hP) ↔ (∀ h, h0, h1. h0 ⊆ h ∧ h1 ⊆ h ∧ hP(h0) ∧ hP(h1) → h0 = h1)

• Proof step 4:

  ∃ h, h0, h1. ¬(h0 ⊆ h ∧ h1 ⊆ h ∧ tr(r, t)(h0) ∧ tr(r, t)(h1) → h0 = h1) ⊦

Rule argument

Additional information

Used terms: tr(r, t)

Proof step 3 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 4 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, tr(r, t)(h0), tr(r, t)(h1) ⊦

Rule argument

Additional information

Proof step 5 from proof precise-tr (previous step) (back to tree)

User applied rule structural induction

And obtained the following new goals

• Proof step 6:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, tr(r, branch(t0, a, t1))(h1), tr(r, branch(t0, a, t1))(h0), ∀ h0, h1, h, r. ¬(h0 ≠ h1 ∧ h0 ⊆ h ∧ h1 ⊆ h ∧ tr(r, t1)(h0) ∧ tr(r, t1)(h1)), ∀ h0, h1, h, r. ¬(h0 ≠ h1 ∧ h0 ⊆ h ∧ h1 ⊆ h ∧ tr(r, t0)(h0) ∧ tr(r, t0)(h1)) ⊦

Rule argument

Additional information

Used simplifier rules: tr-base: ⊦ tr(r, leaf) = (emp ∧ ⌜r = null⌝ ); Tree[A] freely generated by leaf, branch; ⊦ (hP0 ∧ hP1)(h) ↔ hP0(h) ∧ hP1(h); (s) ⊦ ⌜bv⌝ (h) ↔ bv; (s) ⊦ ⌜true⌝ = true; (s) ⊦ (true ∧ hP) = hP; (s) and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep ⊦ emp(h) ↔ h = ∅; (s)

Proof step 6 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 7:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, tr(r, branch(t0, a, t1))(h0), tr(r, branch(t0, a, t1))(h1), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

Proof step 7 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 8:

  ⊦ tr(r, branch(t0, a, t1)) = _∃(λ r0, r1. r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))

• Proof step 9:

  h0 ≠ h1, (_∃(λ r0, r1. r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1)))(h1), (_∃(λ r0, r1. r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1)))(h0), h0 ⊆ h, h1 ⊆ h, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used terms: r, t0, a, t1

Proof step 8 from proof precise-tr (previous step) (back to tree)

Lemma tr-rec

Proof step 9 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, (r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))(h0), (r => mknode(r2, a, r3) * tr(r2, t0) * tr(r3, t1))(h1), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ (_∃ bcP)(h) ↔ (∃ b, c. bcP(b, c)(h)); (s)

Proof step 10 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 11:

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

• Proof step 12:

  h0 ≠ h1, h0 ⊆ h, h1 ⊆ h, (r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))(h0), ∃ h0, h2. h0 ⊥ h2 ∧ h1 = h0 ∪ h2 ∧ (r => mknode(r2, a, r3))(h0) ∧ (tr(r2, t0) * tr(r3, t1))(h2), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used simplifier rules: sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep Used terms: h1, r => mknode(r2, a, r3), tr(r2, t0) * tr(r3, t1)

Proof step 11 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 12 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  h0 ≠ h2 ∪ h3, h0 ⊆ h, h2 ⊥ h3, h2 ∪ h3 ⊆ h, (r => mknode(r2, a, r3))(h2), (r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))(h0), (tr(r2, t0) * tr(r3, t1))(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

Proof step 13 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 14:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

• Proof step 15:

  h2 = (∅ ++ r)(r := mknode(r2, a, r3)), h0 ≠ h2 ∪ h3, r ≠ null, h0 ⊆ h, h2 ∪ h3 ⊆ h, h2 ⊥ h3, (r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))(h0), (tr(r2, t0) * tr(r3, t1))(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used terms: r, mknode(r2, a, r3), h2

Proof step 14 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 15 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 16:

  h0 ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (r => mknode(r0, a, r1) * tr(r0, t0) * tr(r1, t1))(h0), (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

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

Proof step 16 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 17:

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

• Proof step 18:

  h0 ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ∃ h1, h2. h1 ⊥ h2 ∧ h0 = h1 ∪ h2 ∧ (r => mknode(r0, a, r1))(h1) ∧ (tr(r0, t0) * tr(r1, t1))(h2), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used simplifier rules: sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep Used terms: h0, r => mknode(r0, a, r1), tr(r0, t0) * tr(r1, t1)

Proof step 17 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 18 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  h1 ∪ h2 ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, h1 ⊥ h2, h1 ∪ h2 ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (r => mknode(r0, a, r1))(h1), (tr(r0, t0) * tr(r1, t1))(h2), (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

Proof step 19 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 20:

  ⊦ (r => o)(h) ↔ h = (∅ ++ r)(r := o) ∧ r ≠ null

• Proof step 21:

  h1 = (∅ ++ r)(r := mknode(r0, a, r1)), h1 ∪ h2 ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, h1 ∪ h2 ⊆ h, h1 ⊥ h2, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (tr(r2, t0) * tr(r3, t1))(h3), (tr(r0, t0) * tr(r1, t1))(h2), ¬ r ∈ h3, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used terms: r, mknode(r0, a, r1), h1

Proof step 20 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 21 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 22:

  (h2 ++ r)(r := mknode(r0, a, r1)) ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, (h2 ++ r)(r := mknode(r0, a, r1)) ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (tr(r0, t0) * tr(r1, t1))(h2), (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ¬ r ∈ h2, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

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

Proof step 22 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 23:

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

• Proof step 24:

  (h2 ++ r)(r := mknode(r0, a, r1)) ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, (h2 ++ r)(r := mknode(r0, a, r1)) ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ¬ r ∈ h2, ∃ h0, h1. h0 ⊥ h1 ∧ h2 = h0 ∪ h1 ∧ tr(r1, t1)(h0) ∧ tr(r0, t0)(h1), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used terms: h2, tr(r1, t1), tr(r0, t0)

Proof step 23 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 24 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 25:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r1, t1)(h0), tr(r0, t0)(h1), (tr(r2, t0) * tr(r3, t1))(h3), ¬ r ∈ h3, ¬ r ∈ h0, ¬ r ∈ h1, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

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

Proof step 25 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 26:

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

• Proof step 27:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ (h3 ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h0 ⊥ h1, (h3 ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r0, t0)(h1), tr(r1, t1)(h0), ¬ r ∈ h3, ¬ r ∈ h0, ¬ r ∈ h1, ∃ h0, h1. h0 ⊥ h1 ∧ h3 = h0 ∪ h1 ∧ tr(r3, t1)(h0) ∧ tr(r2, t0)(h1), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

Rule argument

Additional information

Used terms: h3, tr(r3, t1), tr(r2, t0)

Proof step 26 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 27 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 28:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h2 ⊥ h4, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r1, t1)(h0), tr(r0, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

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

Proof step 28 from proof precise-tr (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 29:

  r0 = r2, r1 = r3, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h0 ⊥ h1, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, tr(r0, t0)(h1), tr(r1, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦

• Proof step 39:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h0 ⊥ h1, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, tr(r0, t0)(h1), tr(r1, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used formulas: r0 = r2 ∧ r1 = r3

Proof step 29 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 30:

  ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r3, t1)(h0), tr(r2, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3) ⊦

Rule argument

Additional information

Proof step 30 from proof precise-tr (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 31:

  ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h0 ⊥ h1, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, tr(r2, t0)(h1), tr(r3, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), h4 = h1 ∨ ¬ h4 ⊆ h ∨ ¬ h1 ⊆ h ∨ ¬ tr(r2, t0)(h4) ∨ ¬ tr(r2, t0)(h1) ⊦

Rule argument

Additional information

Used terms: h4, h1, h, r2

Proof step 31 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 32:

  ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r3, t1)(h0), tr(r2, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), h4 = h1 ∨ ¬ h4 ⊆ h ∨ ¬ h1 ⊆ h ⊦

Rule argument

Additional information

Proof step 32 from proof precise-tr (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 33:

  ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h0 ⊥ h1, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, tr(r2, t0)(h1), tr(r3, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, h2 = h0 ∨ ¬ h2 ⊆ h ∨ ¬ h0 ⊆ h ∨ ¬ tr(r3, t1)(h2) ∨ ¬ tr(r3, t1)(h0), h4 = h1 ∨ ¬ h4 ⊆ h ∨ ¬ h1 ⊆ h ⊦

Rule argument

Additional information

Used terms: h2, h0, h, r3

Proof step 33 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 34:

  ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) ⊆ h, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r3, t1)(h0), tr(r2, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, h2 = h0 ∨ ¬ h2 ⊆ h ∨ ¬ h0 ⊆ h, h4 = h1 ∨ ¬ h4 ⊆ h ∨ ¬ h1 ⊆ h ⊦

Rule argument

Additional information

Proof step 34 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 35:

  ⊦ r ≠ null ∧ ¬ r ∈ h0 → ((h0 ++ r)(r := d) ⊆ h ↔ h0 ⊆ h ∧ r ∈ h ∧ h(r) = d)

• Proof step 36:

  h(r) = mknode(r2, a, r3), ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h0 ⊥ h1, h2 ∪ h4 ⊆ h, h2 ⊥ h4, r ∈ h, tr(r2, t0)(h1), tr(r3, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h2 ∪ h4, ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, h2 = h0 ∨ ¬ h2 ⊆ h ∨ ¬ h0 ⊆ h, h4 = h1 ∨ ¬ h4 ⊆ h ∨ ¬ h1 ⊆ h ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) Used terms: h2 ∪ h4, r, mknode(r2, a, r3), h

Proof step 35 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 36 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 37:

  h(r) = mknode(r2, a, r3), ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r2, a, r3)) ⊆ h, h2 ⊥ h4, h2 ⊆ h, h4 ⊆ h, r ∈ h, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r3, t1)(h0), tr(r2, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, h2 = h0 ∨ ¬ h0 ⊆ h, h4 = h1 ∨ ¬ h1 ⊆ h ⊦

Rule argument

Additional information

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

Proof step 37 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 38:

  ⊦ r ≠ null ∧ ¬ r ∈ h0 → ((h0 ++ r)(r := d) ⊆ h ↔ h0 ⊆ h ∧ r ∈ h ∧ h(r) = d)

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s) ⊦ h0 ⊥ h1 → (h0 ∪ h1 ⊆ h ↔ h0 ⊆ h ∧ h1 ⊆ h); (ws) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) Used terms: h0 ∪ h1, r, mknode(r2, a, r3), h

Proof step 38 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 39 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 40:

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

• Proof step 41:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h2 ⊥ h4, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r1, t1)(h0), tr(r0, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ r0. (r0 ∈ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) → r0 ∈ h) ∧ (r0 ∈ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)) → ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3))(r0) = h(r0)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used terms: ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), h

Proof step 40 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 41 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 42:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ⊆ h, h0 ⊥ h1, h2 ⊥ h4, tr(r0, t0)(h1), tr(r1, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ r4. (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → (r4 = r ⊃ mknode(r2, a, r3);((h2 ∪ h4) ++ r)(r4)) = h(r4)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

Proof step 42 from proof precise-tr (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 43:

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

• Proof step 44:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, h2 ⊥ h4, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r1, t1)(h0), tr(r0, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ r2. (r2 ∈ ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) → r2 ∈ h) ∧ (r2 ∈ ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) → ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1))(r2) = h(r2)), ∀ r4. (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → (r4 = r ⊃ mknode(r2, a, r3);((h2 ∪ h4) ++ r)(r4)) = h(r4)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used terms: ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)), h

Proof step 43 from proof precise-tr (previous step) (back to tree)

A spec-lemma

Proof step 44 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 45:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, h2 ⊥ h4, tr(r0, t0)(h1), tr(r1, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ r4. (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → (r4 = r ⊃ mknode(r2, a, r3);((h2 ∪ h4) ++ r)(r4)) = h(r4)), ∀ r4. (r4 = r ∨ r4 ∈ h0 ∨ r4 ∈ h1 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h0 ∨ r4 ∈ h1 → (r4 = r ⊃ mknode(r0, a, r1);((h0 ∪ h1) ++ r)(r4)) = h(r4)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used simplifier rules: r0 ≠ null, ¬ r0 ∈ h ⊦ r ∈ h ++ r0 ↔ r = r0 ∨ r ∈ h; (s) ⊦ h0 ⊥ h1 → (r ∈ h0 ∪ h1 ↔ r ∈ h0 ∨ r ∈ h1); (s) ⊦ h(r0 := o)(r) = (r = r0 ∧ r0 ∈ h ⊃ o;h(r)); (s) ⊦ r ∈ h(r0 := o) ↔ r ∈ h; (s)

Proof step 45 from proof precise-tr (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 46:

  ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), r ≠ null, h0 ⊥ h1, h2 ⊥ h4, tr(r3, t1)(h2), tr(r2, t0)(h4), tr(r1, t1)(h0), tr(r0, t0)(h1), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), r = r ∨ r ∈ h0 ∨ r ∈ h1 → r ∈ h, r = r ∨ r ∈ h0 ∨ r ∈ h1 → (r = r ⊃ mknode(r0, a, r1);((h0 ∪ h1) ++ r)(r)) = h(r), ∀ r4. (r4 = r ∨ r4 ∈ h0 ∨ r4 ∈ h1 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h0 ∨ r4 ∈ h1 → (r4 = r ⊃ mknode(r0, a, r1);((h0 ∪ h1) ++ r)(r4)) = h(r4)), ∀ r4. (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → (r4 = r ⊃ mknode(r2, a, r3);((h2 ∪ h4) ++ r)(r4)) = h(r4)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used terms: r

Proof step 46 from proof precise-tr (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 47:

  mknode(r0, a, r1) = h(r), ((h0 ∪ h1) ++ r)(r := mknode(r0, a, r1)) ≠ ((h2 ∪ h4) ++ r)(r := mknode(r2, a, r3)), h0 ⊥ h1, h2 ⊥ h4, r ∈ h, tr(r0, t0)(h1), tr(r1, t1)(h0), tr(r2, t0)(h4), tr(r3, t1)(h2), ¬ r ∈ h0, ¬ r ∈ h1, ¬ r ∈ h2, ¬ r ∈ h4, ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t1)(h2) ∨ ¬ tr(r0, t1)(h3), ∀ h2, h3, h4, r0. h2 = h3 ∨ ¬ h2 ⊆ h4 ∨ ¬ h3 ⊆ h4 ∨ ¬ tr(r0, t0)(h2) ∨ ¬ tr(r0, t0)(h3), ∀ r4. (r4 ∈ h2 ∨ r4 ∈ h4 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h2 ∨ r4 ∈ h4 → (r4 = r ⊃ mknode(r2, a, r3);((h2 ∪ h4) ++ r)(r4)) = h(r4)), ∀ r4. (r4 ∈ h0 ∨ r4 ∈ h1 → r4 ∈ h) ∧ (r4 = r ∨ r4 ∈ h0 ∨ r4 ∈ h1 → (r4 = r ⊃ mknode(r0, a, r1);((h0 ∪ h1) ++ r)(r4)) = h(r4)) ⊦ r0 = r2 ∧ r1 = r3

Rule argument

Additional information

Used simplifier rules: ⊦ r ∈ h → r ≠ null; (s)

Proof step 47 from proof precise-tr (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ mknode(left, val, left0) = mknode(left1, val0, left2) ↔ left = left1 ∧ val = val0 ∧ left0 = left2; (ws) Used terms: r