Proof step 1 from proof Sep-forall (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦

Rule argument

Additional information

Proof step 2 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h) ⊦

Rule argument

Additional information

Proof step 3 from proof Sep-forall (previous step) (back to tree)

User applied rule induction

And obtained the following new goals

• Proof step 4:

  (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h), ∀ st0, adP0, h, adP. # st0 < # st → (adP0 ⊗_ st0)(h) ∧ ¬ (adP ⊗_ st0)(h) ∧ (∀ a, h. a ∈ st0 ∧ adP0(a, st0(a))(h) → adP(a, st0(a))(h)) → false ⊦

Rule argument

Additional information

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

Proof step 4 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h), ∀ st0, adP0, h, adP. # st0 < # st → (adP0 ⊗_ st0)(h) ∧ ¬ (adP ⊗_ st0)(h) ∧ (∀ a, h. a ∈ st0 ∧ adP0(a, st0(a))(h) → adP(a, st0(a))(h)) → false ⊦

Rule argument

Additional information

Proof step 5 from proof Sep-forall (previous step) (back to tree)

User applied rule cut formula

And obtained the following new goals

• Proof step 6:

  st ≠ ∅, (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h), ∀ st0, adP0, h, adP. # st0 < # st → (adP0 ⊗_ st0)(h) ∧ ¬ (adP ⊗_ st0)(h) ∧ (∀ a, h. a ∈ st0 ∧ adP0(a, st0(a))(h) → adP(a, st0(a))(h)) → false ⊦

Rule argument

Additional information

Used simplifier rules: Sep-emp: ⊦ adP ⊗_ ∅ = emp; ⊦ emp(h) ↔ h = ∅; (s) Used formulas: st = ∅

Proof step 6 from proof Sep-forall (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 7:

  ⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0(a := d) ∧ ¬ a ∈ st0)

• Proof step 8:

  (adP0 ⊗_ st)(h), ¬ (adP ⊗_ st)(h), ∀ a, h. a ∈ st ∧ adP0(a, st(a))(h) → adP(a, st(a))(h), ∃ a, d, st0. st = st0(a := d) ∧ ¬ a ∈ st0, ∀ st0, adP0, h, adP. # st0 < # st → (adP0 ⊗_ st0)(h) ∧ ¬ (adP ⊗_ st0)(h) ∧ (∀ a, h. a ∈ st0 ∧ adP0(a, st0(a))(h) → adP(a, st0(a))(h)) → false ⊦

Rule argument

Additional information

Used terms: st

Proof step 7 from proof Sep-forall (previous step) (back to tree)

A spec-lemma

Proof step 8 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  (adP0(a, d) * adP0 ⊗_ st0)(h), ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦

Rule argument

Additional information

Used simplifier rules: Sep-ins: ⊦ ¬ a ∈ st → adP ⊗_ st(a := d) = adP(a, d) * adP ⊗_ st; ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s)

Proof step 9 from proof Sep-forall (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 10:

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

• Proof step 11:

  ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∃ h0, h1. h0 ⊥ h1 ∧ h = h0 ∪ h1 ∧ adP0(a, d)(h0) ∧ (adP0 ⊗_ st0)(h1), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦

Rule argument

Additional information

Used terms: h, adP0 ⊗_ st0, adP0(a, d)

Proof step 10 from proof Sep-forall (previous step) (back to tree)

A spec-lemma

Proof step 11 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 12:

  adP0(a, d)(h0), (adP0 ⊗_ st0)(h1), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦

Rule argument

Additional information

Proof step 12 from proof Sep-forall (previous step) (back to tree)

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:

  (adP0 ⊗_ st0)(h1), adP0(a, d)(h0), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦ adP(a, d)(h0) ∧ (adP ⊗_ st0)(h1) ∧ h0 ⊥ h1

Rule argument

Additional information

Used terms: h1, h0, adP ⊗_ st0, adP(a, d)

Proof step 13 from proof Sep-forall (previous step) (back to tree)

A spec-lemma

Proof step 14 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 15:

  adP0(a, d)(h0), (adP0 ⊗_ st0)(h1), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦ adP(a, d)(h0) ∧ (adP ⊗_ st0)(h1)

Rule argument

Additional information

Proof step 15 from proof Sep-forall (previous step) (back to tree)

User applied rule apply induction

And obtained the following new goals

• Proof step 16:

  (adP0 ⊗_ st0)(h1), adP0(a, d)(h0), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦ adP(a, d)(h0) ∧ (adP ⊗_ st0)(h1), (adP0 ⊗_ st0)(h1) ∧ ¬ (adP ⊗_ st0)(h1) ∧ (∀ a, h. a ∈ st0 ∧ adP0(a, st0(a))(h) → adP(a, st0(a))(h))

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → # st(a := d) = # st + 1; (s) ⊦ 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: st0, adP0, h1, adP

Proof step 16 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 17:

  adP0(a, d)(h0), (adP0 ⊗_ st0)(h1), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦ adP(a, d)(h0) ∧ (adP ⊗_ st0)(h1), ¬ (adP ⊗_ st0)(h1) ∧ (a0 ∈ st0 ∧ adP0(a0, st0(a0))(h) → adP(a0, st0(a0))(h))

Rule argument

Additional information

Proof step 17 from proof Sep-forall (previous step) (back to tree)

User applied rule all left

And obtained the following new goals

• Proof step 18:

  (adP0 ⊗_ st0)(h1), adP0(a, d)(h0), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), (a = a ∨ a ∈ st0) ∧ adP0(a, st0(a := d)(a))(h0) → adP(a, st0(a := d)(a))(h0), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦ adP(a, d)(h0) ∧ (adP ⊗_ st0)(h1), ¬ (adP ⊗_ st0)(h1) ∧ (a0 ∈ st0 ∧ adP0(a0, st0(a0))(h) → adP(a0, st0(a0))(h))

Rule argument

Additional information

Used terms: a, h0

Proof step 18 from proof Sep-forall (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 19:

  a0 ∈ st0, adP(a, d)(h0), adP0(a, d)(h0), (adP0 ⊗_ st0)(h1), adP0(a0, st0(a0))(h), h0 ⊥ h1, ¬ a ∈ st0, ¬ (adP(a, d) * adP ⊗_ st0)(h0 ∪ h1), ¬ (adP ⊗_ st0)(h1), ¬ adP(a0, st0(a0))(h), ∀ a0, h0. (a0 = a ∨ a0 ∈ st0) ∧ adP0(a0, st0(a := d)(a0))(h0) → adP(a0, st0(a := d)(a0))(h0), ∀ st1, adP0, h, adP. # st1 < # st0(a := d) → (adP0 ⊗_ st1)(h) ∧ ¬ (adP ⊗_ st1)(h) ∧ (∀ a, h. a ∈ st1 ∧ adP0(a, st1(a))(h) → adP(a, st1(a))(h)) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ a = b → st(b := d)(a) = d; (s)

Proof step 19 from proof Sep-forall (previous step) (back to tree)

User applied rule all left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ b ∈ st ∧ ¬ a ∈ st → st(a := d)(b) = st(b); (s) Used terms: a0, h