Proof step 1 from proof And-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 And-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 And-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 And-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 And-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: And-emp: ⊦ adP ∧_ ∅ = true; ⊦ true(h); (s) Used formulas: st = ∅

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

A spec-lemma

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 9:

  adP0(a, d)(h), (adP0 ∧_ st0)(h), ¬ a ∈ st0, ∀ 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)(h) ∧ (adP ∧_ st0)(h)

Rule argument

Additional information

Used simplifier rules: And-ins: ⊦ ¬ a ∈ st → adP ∧_ st(a := d) = (adP(a, d) ∧ adP ∧_ st); ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s) and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep ⊦ (hP0 ∧ hP1)(h) ↔ hP0(h) ∧ hP1(h); (s)

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

User applied rule all left

And obtained the following new goals

• Proof step 10:

  (adP0 ∧_ st0)(h), adP0(a, d)(h), ¬ a ∈ st0, (a = a ∨ a ∈ st0) ∧ adP0(a, st0(a := d)(a))(h) → adP(a, st0(a := d)(a))(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 ⊦ adP(a, d)(h) ∧ (adP ∧_ st0)(h)

Rule argument

Additional information

Used terms: a, h

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 11:

  adP(a, d)(h), adP0(a, d)(h), (adP0 ∧_ st0)(h), ¬ a ∈ st0, ¬ (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: ⊦ a = b → st(b := d)(a) = d; (s)

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

User applied rule apply induction

And obtained the following new goals

• Proof step 12:

  adP(a, d)(h), (adP0 ∧_ st0)(h), adP0(a, d)(h), ¬ a ∈ st0, ¬ (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 ⊦ (adP0 ∧_ st0)(h) ∧ ¬ (adP ∧_ st0)(h) ∧ (∀ 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, h, adP

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 13:

  a0 ∈ st0, adP(a, d)(h), adP0(a, d)(h), (adP0 ∧_ st0)(h), adP0(a0, st0(a0))(h0), ¬ a ∈ st0, ¬ (adP ∧_ st0)(h), ¬ adP(a0, st0(a0))(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 ⊦

Rule argument

Additional information

Proof step 13 from proof And-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, h0