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Proof step 3 from proof inex (previous step)

a ∈ st ⊦ ∃ d, st0. (∀ a0. (a0 ∈ st ↔ a0 ∈ st0[a, d]) ∧ st[a0] = st0[a, d][a0]) ∧ ¬ a ∈ st0

User applied rule exists right

And obtained the following new goals

Proof step 4:
a ∈ st ⊦ (∀ a0. (a0 ∈ st ↔ a0 ∈ (st -- a)[a, st[a]]) ∧ st[a0] = (st -- a)[a, st[a]][a0]) ∧ ¬ a ∈ st -- a

Rule argument

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

Additional information

Used terms: st[a], st -- a