Proof step 1 from proof alloc (conclusion) (back to tree)

⊦ r ≠ null ∧ ¬ r ∈ rs → (hR[rs, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)])

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬(hR[rs, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 2 from proof alloc (previous step) (back to tree)

¬(hR[rs, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
¬(hR[rs, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 3 from proof alloc (previous step) (back to tree)

¬(hR[rs, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 4:
¬ r ∈ rs, r ≠ null ⊦ hR[rs, f, g] ↔ (r ==>_ * hR)[rs ++ r, f, g]
Proof step 5:
¬((r ==>_ * hR)[rs ++ r, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Rule argument

spec: heap-aux, name:mapex-new, seq:¬ r ∈ rs, r ≠ null ⊦ hR[rs, f, g] ↔ (r ==>_ * hR)[rs ++ r, f, g], subst:[hR, r, rs, g, f] → [hR, r, rs, g, f], <all paths>

Additional information

Used terms: hR, r, rs, g, f

Proof step 4 from proof alloc (previous step) (back to tree)

¬ r ∈ rs, r ≠ null ⊦ hR[rs, f, g] ↔ (r ==>_ * hR)[rs ++ r, f, g]

A spec-lemma

Proof step 5 from proof alloc (previous step) (back to tree)

¬((r ==>_ * hR)[rs ++ r, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
¬((r ==>_ * hR)[rs ++ r, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 6 from proof alloc (previous step) (back to tree)

¬((r ==>_ * hR)[rs ++ r, f, g] ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]), r ≠ null, ¬ r ∈ rs ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ (r ==>_ * hR)[rs, f, g] ↔ (∃ a. (a[r =>_] * hR)[rs, f, g])
Proof step 8:
r ≠ null, ¬ r ∈ rs ⊦ (∃ a. (a[r =>_] * hR)[rs ++ r, f, g]) ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]

Rule argument

spec: <toplevel>, name:lift-unknown, seq: ⊦ (r ==>_ * hR)[rs, f, g] ↔ (∃ a. (a[r =>_] * hR)[rs, f, g]), subst:[hR, r, rs, g, f] → [hR, r, rs ++ r, g, f], <all paths>

Additional information

Used terms: hR, r, rs ++ r, g, f

Proof step 7 from proof alloc (previous step) (back to tree)

⊦ (r ==>_ * hR)[rs, f, g] ↔ (∃ a. (a[r =>_] * hR)[rs, f, g])

Lemma lift-unknown

Proof step 8 from proof alloc (previous step) (back to tree)

r ≠ null, ¬ r ∈ rs ⊦ (∃ a. (a[r =>_] * hR)[rs ++ r, f, g]) ↔ (a[r =>_] * hR)[rs ++ r, f, g(r; a)]

User applied rule case distinction

And obtained the following new goals

Proof step 9:
r ≠ null, ¬ (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs, ∃ a. (a[r =>_] * hR)[rs ++ r, f, g] ⊦
Proof step 12:
r ≠ null, (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs ⊦ ∃ a. (a[r =>_] * hR)[rs ++ r, f, g]

Rule argument

pos: right 1

Additional information

Proof step 9 from proof alloc (previous step) (back to tree)

r ≠ null, ¬ (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs, ∃ a. (a[r =>_] * hR)[rs ++ r, f, g] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
r ≠ null, (a0[r =>_] * hR)[rs ++ r, f, g], ¬ (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs ⊦

Rule argument

Additional information

Proof step 10 from proof alloc (previous step) (back to tree)

r ≠ null, (a0[r =>_] * hR)[rs ++ r, f, g], ¬ (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0

Rule argument

spec: <toplevel>, name:aux-unknown, seq:⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0, subst:[hR, a0, r, rs, a1, g, f] → [hR, a0, r, rs ++ r, a, g, f], <all paths>

Additional information

Used terms: hR, a0, r, rs ++ r, a, g, f

Proof step 11 from proof alloc (previous step) (back to tree)

⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0

Lemma aux-unknown

Proof step 12 from proof alloc (previous step) (back to tree)

r ≠ null, (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ r ∈ rs ⊦ ∃ a. (a[r =>_] * hR)[rs ++ r, f, g]

User applied rule exists right

And obtained the following new goals

Proof step 13:
r ≠ null, (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ (g(r)[r =>_] * hR)[rs ++ r, f, g], ¬ r ∈ rs ⊦

Rule argument

pos: right 1 terms: g(r) abort: F discard: T computed: F

Additional information

Used terms: g(r)

Proof step 13 from proof alloc (previous step) (back to tree)

r ≠ null, (a[r =>_] * hR)[rs ++ r, f, g(r; a)], ¬ (g(r)[r =>_] * hR)[rs ++ r, f, g], ¬ r ∈ rs ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0

Rule argument

spec: <toplevel>, name:aux-unknown, seq:⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0, subst:[hR, a0, r, rs, a1, g, f] → [hR, g(r), r, rs ++ r, a, g, f], <all paths>

Additional information

Used terms: hR, g(r), r, rs ++ r, a, g, f

Proof step 14 from proof alloc (previous step) (back to tree)

⊦ (a0[r =>_] * hR)[rs, f, g] ↔ (a1[r =>_] * hR)[rs, f, g(r; a1)] ∧ g(r) = a0

Lemma aux-unknown

Proof Tree for alloc (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.