Proof step 1 from proof map-putex (conclusion) (back to tree)

r ∈ h ⊦ (r => o0 * hP)(h(r := o1)) ↔ o0 = o1 ∧ (r.=>_ * hP)(h)

System applied rule normalize

And obtained the following new goals

Proof step 2:
¬((r => o0 * hP)(h(r := o1)) ↔ o0 = o1 ∧ (r.=>_ * hP)(h)), r ∈ h ⊦

Rule argument

Additional information

Proof step 2 from proof map-putex (previous step) (back to tree)

¬((r => o0 * hP)(h(r := o1)) ↔ o0 = o1 ∧ (r.=>_ * hP)(h)), r ∈ h ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 3:
r ∈ h, (r => o0 * hP)(h(r := o1)) ⊦ o0 = o1 ∧ (r.=>_ * hP)(h)

Rule argument

pos: left 1

Additional information

Used simplifier rules: mapex: ⊦ r.=>_ = _∃(λ o. r => o); ⊦ (_∃ bP * hP)(h) ↔ (∃ b. (bP(b) * hP)(h)); (s) ⊦ (r => o * hP)(h) → r ∈ h; (ss) r ∈ h ⊦ (r => o * hP)(h) → ((r => o0 * hP)(h(r := o1)) ↔ o0 = o1); (ss) sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep

Proof step 3 from proof map-putex (previous step) (back to tree)

r ∈ h, (r => o0 * hP)(h(r := o1)) ⊦ o0 = o1 ∧ (r.=>_ * hP)(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r ∈ h, (r => o1 * hP)(h(r := o1)) ⊦ ∃ b. (r => b * hP)(h)

Rule argument

Additional information

Used simplifier rules: mapex: ⊦ r.=>_ = _∃(λ o. r => o); sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep map-write: r ∈ h ⊦ (r => o0 * hP)(h(r := o1)) → o0 = o1; (forward) from specification refmap-maplet ⊦ (_∃ bP * hP)(h) ↔ (∃ b. (bP(b) * hP)(h)); (s)

Proof step 4 from proof map-putex (previous step) (back to tree)

r ∈ h, (r => o1 * hP)(h(r := o1)) ⊦ ∃ b. (r => b * hP)(h)

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ st ↔ (∃ d, st0. st = st0(a := d) ∧ ¬ a ∈ st0)
Proof step 6:
(r => o1 * hP)(h(r := o1)), ∃ d, st0. h = st0(r := d) ∧ ¬ r ∈ st0 ⊦ ∃ b. (r => b * hP)(h)

Rule argument

spec: map, name:exin, seq: ⊦ a ∈ st ↔ (∃ d, st0. st = st0(a := d) ∧ ¬ a ∈ st0), subst:[a, st] → [r, h]; [A, B] → [Ref[A], A], lhs:r ∈ h, <all paths>

Additional information

Used terms: r, h

Proof step 5 from proof map-putex (previous step) (back to tree)

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

A spec-lemma

Proof step 6 from proof map-putex (previous step) (back to tree)

(r => o1 * hP)(h(r := o1)), ∃ d, st0. h = st0(r := d) ∧ ¬ r ∈ st0 ⊦ ∃ b. (r => b * hP)(h)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(r => o1 * hP)(h0(r := o1)), ¬ r ∈ h0 ⊦ ∃ b. (r => b * hP)(h0(r := o))

Rule argument

Additional information

Used simplifier rules: ⊦ a = b → st(a := d)(b := d0) = st(b := d0); (s)

Proof step 7 from proof map-putex (previous step) (back to tree)

(r => o1 * hP)(h0(r := o1)), ¬ r ∈ h0 ⊦ ∃ b. (r => b * hP)(h0(r := o))

User applied rule exists right

And obtained the following new goals

Proof step 8:
(r => o1 * hP)(h0(r := o1)), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

Rule argument

pos: right 1 terms: o abort: F discard: T computed: F

Additional information

Used terms: o

Proof step 8 from proof map-putex (previous step) (back to tree)

(r => o1 * hP)(h0(r := o1)), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null
Proof step 10:
h0(r := o1)(r) = o1, r ≠ null, hP(h0(r := o1) -- r), r ∈ h0(r := o1), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

Rule argument

spec: refmap-maplet, name:map-del, seq: ⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null, subst:[h, r, o, hP] → [h0(r := o1), r, o1, hP]; [A] → [A], lhs:(r => o1 * hP)(h0(r := o1)), <all paths>

Additional information

Used simplifier rules: sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep Used terms: h0(r := o1), r, o1, hP

Proof step 9 from proof map-putex (previous step) (back to tree)

⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null

A spec-lemma

Proof step 10 from proof map-putex (previous step) (back to tree)

h0(r := o1)(r) = o1, r ≠ null, hP(h0(r := o1) -- r), r ∈ h0(r := o1), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
r ≠ null, hP(h0), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → st -- a = st; (s) ⊦ st(a := d) -- a = st -- a; (s) ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s) ⊦ a = b → st(b := d)(a) = d; (s)

Proof step 11 from proof map-putex (previous step) (back to tree)

r ≠ null, hP(h0), ¬ (r => o * hP)(h0(r := o)), ¬ r ∈ h0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null

Rule argument

spec: refmap-maplet, name:map-del, seq: ⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null, subst:[h, r, o, hP] → [h0(r := o), r, o, hP]; [A] → [A], lhs:(r => o * hP)(h0(r := o)), <all paths>

Additional information

Used simplifier rules: ⊦ st(a := d) -- a = st -- a; (s) ⊦ ¬ a ∈ st → st -- a = st; (s) ⊦ a ∈ st(b := d) ↔ a = b ∨ a ∈ st; (s) ⊦ a = b → st(b := d)(a) = d; (s) sep-ass: ⊦ (hP0 * hP1) * hP2 = hP0 * hP1 * hP2; from specification map-sep sep-com: ⊦ hP0 * hP1 = hP1 * hP0; from specification map-sep Used terms: h0(r := o), r, o, hP

Proof step 12 from proof map-putex (previous step) (back to tree)

⊦ (r => o * hP)(h) ↔ hP(h -- r) ∧ r ∈ h ∧ h(r) = o ∧ r ≠ null

A spec-lemma

Proof Tree for map-putex (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.