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

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

User applied rule apply lemma

And obtained the following new goals

Proof step 2:
⊦ a ∈ st ↔ (∃ d, st0. st = st0[a, d] ∧ ¬ a ∈ st0)
Proof step 3:
a ∈ st[a, d] ↔ (∃ d0, st0. st[a, d] = st0[a, d0] ∧ ¬ a ∈ st0) ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Rule argument

spec: <toplevel>, name:exin, seq: ⊦ a ∈ st ↔ (∃ d, st0. st = st0[a, d] ∧ ¬ a ∈ st0), subst:[a, st] → [a, st[a, d]]

Additional information

Used terms: a, st[a, d]

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

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

Lemma exin

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

a ∈ st[a, d] ↔ (∃ d0, st0. st[a, d] = st0[a, d0] ∧ ¬ a ∈ st0) ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
st[a, d] = st0[a, d0], ¬ a ∈ st0 ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Rule argument

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st;

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

st[a, d] = st0[a, d0], ¬ a ∈ st0 ⊦ ∃ st0. st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

User applied rule exists right

And obtained the following new goals

Proof step 5:
st[a, d] = st0[a, d0], ¬ a ∈ st0 ⊦ st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Rule argument

pos: right 1 terms: st0 abort: F discard: T computed: T

Additional information

Used terms: st0

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

st[a, d] = st0[a, d0], ¬ a ∈ st0 ⊦ st0[a, d] = st[a, d] ∧ ¬ a ∈ st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
st[a, d] = st0[a, d0], st0[a, d] ≠ st0[a, d0], ¬ a ∈ st0 ⊦

Rule argument

Additional information

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

st[a, d] = st0[a, d0], st0[a, d] ≠ st0[a, d0], ¬ a ∈ st0 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ d ≠ d0 → st[a, d] ≠ st0[a, d0]

Rule argument

spec: <toplevel>, name:put-neq, seq: ⊦ d ≠ d0 → st[a, d] ≠ st0[a, d0], subst:[st, d, d0, st0, a] → [st, d, d0, st0, a], <all paths>

Additional information

Used terms: st, d, d0, st0, a

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

⊦ d ≠ d0 → st[a, d] ≠ st0[a, d0]

Lemma put-neq

Proof Tree for exputin (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.