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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st = ∅ ↔ ¬(∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)

Rule argument

Additional information

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

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

User applied rule case distinction

And obtained the following new goals

Proof step 3:
st = ∅, ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0 ⊦
Proof step 6:
st ≠ ∅ ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

Rule argument

pos: right 1

Additional information

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st0[a, d] = ∅, ¬ a ∈ st0 ⊦

Rule argument

Additional information

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

st0[a, d] = ∅, ¬ a ∈ st0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ st[a, d] ≠ ∅

Rule argument

spec: <toplevel>, name:neq, seq: ⊦ st[a, d] ≠ ∅, subst:[d, a, st] → [d, a, st0], <all paths>

Additional information

Used terms: d, a, st0

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

⊦ st[a, d] ≠ ∅

Lemma neq

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

st ≠ ∅ ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])
Proof step 8:
∃ a. ¬((a ∈ st ↔ a ∈ ∅) ∧ st[a] = ∅[a]) ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

Rule argument

spec: <toplevel>, name:Extension, seq: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]), subst:[st2, st1] → [∅, st], <all paths>

Additional information

Used terms: ∅, st

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

⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a])

Lemma Extension

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0, ¬ a ∈ st ∧ st[a] = ∅[a]

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;

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

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

User applied rule cut formula

And obtained the following new goals

Proof step 10:
a ∈ st ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0, ¬ a ∈ st ∧ st[a] = ∅[a]

Rule argument

fma: a ∈ st

Additional information

Used simplifier rules: notmemlookup: ⊦ ¬ a ∈ st → st[a] = ∅[a]; Used formulas: a ∈ st

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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a ∈ st ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

Rule argument

Additional information

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

a ∈ st ⊦ ∃ a, d, st0. st = st0[a, d] ∧ ¬ a ∈ st0

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: a, st[a], st -- a abort: F discard: T computed: F

Additional information

Used simplifier rules: del-put: ⊦ (st -- a)[a, d] = st[a, d]; ins-same-01: ⊦ a ∈ st → st[a, st[a]] = st; Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; Used terms: a, st[a], st -- a

Proof Tree for exnotempty (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.