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Proof step 4 from proof exputin (previous step)

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 2 terms: st0 abort: F discard: T computed: T

Additional information

Used terms: st0