Proof step 1 from proof And-discard (conclusion) (back to tree)

System applied rule normalize

And obtained the following new goals

• Proof step 2:

  st ≠ ∅, (λ a, d. hP) ∧_ st ≠ hP ⊦

Rule argument

Additional information

Proof step 2 from proof And-discard (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 3:

  st ≠ ∅, (λ a, d. hP) ∧_ st ≠ hP ⊦

Rule argument

Additional information

Proof step 3 from proof And-discard (previous step) (back to tree)

User applied rule induction

And obtained the following new goals

• Proof step 4:

  st ≠ ∅, (λ a, d. hP) ∧_ st ≠ hP, ∀ st0, hP. # st0 < # st → st0 ≠ ∅ ∧ (λ a, d. hP) ∧_ st0 ≠ hP → false ⊦

Rule argument

Additional information

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

Proof step 4 from proof And-discard (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 5:

  st ≠ ∅, (λ a, d. hP) ∧_ st ≠ hP, ∀ st0, hP. # st0 < # st → st0 ≠ ∅ ∧ (λ a, d. hP) ∧_ st0 ≠ hP → false ⊦

Rule argument

Additional information

Proof step 5 from proof And-discard (previous step) (back to tree)

User applied rule apply rewrite lemma

And obtained the following new goals

• Proof step 6:

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

• Proof step 7:

  (λ a, d. hP) ∧_ st ≠ hP, ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0, ∀ st0, hP. # st0 < # st → st0 ≠ ∅ ∧ (λ a, d. hP) ∧_ st0 ≠ hP → false ⊦

Rule argument

Additional information

Used terms: st

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

A spec-lemma

Proof step 7 from proof And-discard (previous step) (back to tree)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 8:

  (hP ∧ (λ a, d. hP) ∧_ st0) ≠ hP, ¬ a ∈ st0, ∀ st1, hP. # st1 < # st0[a, d] → st1 ≠ ∅ ∧ (λ a, d. hP) ∧_ st1 ≠ hP → false ⊦

Rule argument

Additional information

Used simplifier rules: And-ins: ⊦ ¬ a ∈ st → adP ∧_ st[a, d] = (adP(a, d) ∧ adP ∧_ st);

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

User applied rule apply induction

And obtained the following new goals

• Proof step 9:

  (hP ∧ (λ a, d. hP) ∧_ st0) ≠ hP, ¬ a ∈ st0, ∀ st1, hP. # st1 < # st0[a, d] → st1 ≠ ∅ ∧ (λ a, d. hP) ∧_ st1 ≠ hP → false ⊦ st0 ≠ ∅ ∧ (λ a, d. hP) ∧_ st0 ≠ hP

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, hP

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

• Proof step 10:

  (λ a, d. hP) ∧_ st0 = hP, (hP ∧ hP) ≠ hP, st0 ≠ ∅, ¬ a ∈ st0, ∀ st1, hP. # st1 < # st0[a, d] → st1 ≠ ∅ ∧ (λ a, d. hP) ∧_ st1 ≠ hP → false ⊦

Rule argument

Additional information

Used simplifier rules: And-emp: ⊦ adP ∧_ ∅ = true; and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep ⊦ (true ∧ hP) = hP; (s)

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

User applied rule expand left

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ ((hP0 ∧ hP1))(h) ↔ hP0(h) ∧ hP1(h); (s) and-ass: ⊦ ((hP0 ∧ hP1) ∧ hP2) = (hP0 ∧ hP1 ∧ hP2); from specification heap-sep and-com: ⊦ (hP0 ∧ hP1) = (hP1 ∧ hP0); from specification heap-sep