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

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

User applied rule case distinction

And obtained the following new goals

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

Rule argument

pos: right 1

Additional information

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

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

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

User applied rule apply lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: a, st

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

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

Lemma inex

Proof Tree for exin (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.