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

⊦ st ≠ ∅ → (((adP ⊗_ st * hP))(h) ↔ (∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h)))

System applied rule normalize

And obtained the following new goals

Proof step 2:
st ≠ ∅ ⊦ ((adP ⊗_ st * hP))(h) ↔ (∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h))

Rule argument

Additional information

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

st ≠ ∅ ⊦ ((adP ⊗_ st * hP))(h) ↔ (∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
st ≠ ∅, ((adP ⊗_ st * hP))(h) ⊦ ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h)

Rule argument

Additional information

Used simplifier rules: Sep-ins: ⊦ ¬ a ∈ st → adP ⊗_ st[a, d] = adP(a, d) * adP ⊗_ st; sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification heap-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification heap-sep

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

st ≠ ∅, ((adP ⊗_ st * hP))(h) ⊦ ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 5:
((adP ⊗_ st * hP))(h), ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0 ⊦ ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h)

Rule argument

spec: store, name:exnotempty, seq: ⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0), subst:[st] → [st], <all paths>

Additional information

Used terms: st

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

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

A spec-lemma

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

((adP ⊗_ st * hP))(h), ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0 ⊦ ∃ a, d, st0. ¬ a ∈ st0 ∧ st = st0[a, d] ∧ ((adP(a, d) * adP ⊗_ st0 * hP))(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
(((adP(a, d) * adP ⊗_ st0) * hP))(h), ¬ a ∈ st0 ⊦ ∃ a0, d0, st1. ¬ a0 ∈ st1 ∧ st0[a, d] = st1[a0, d0] ∧ ((adP(a0, d0) * adP ⊗_ st1 * hP))(h)

Rule argument

Additional information

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

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

(((adP(a, d) * adP ⊗_ st0) * hP))(h), ¬ a ∈ st0 ⊦ ∃ a0, d0, st1. ¬ a0 ∈ st1 ∧ st0[a, d] = st1[a0, d0] ∧ ((adP(a0, d0) * adP ⊗_ st1 * hP))(h)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a, d, st0

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: a, d, st0

Proof Tree for Sep-any (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.