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Proof step 18 from proof del-exists (previous step)

st0 = st[a0, st0[a0]], a0 ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a0 ∈ st, ∃ st1. ¬ a0 ∈ st1 ∧ st0[a0, d] = st1[a0, d]

User applied rule exists right

And obtained the following new goals

Proof step 19:
st0 = st[a0, st0[a0]], a0 ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a0 ∈ st, ¬ a0 ∈ st ∧ st0[a0, d] = st[a0, d], ∃ st1. ¬ a0 ∈ st1 ∧ st0[a0, d] = st1[a0, d]

Rule argument

pos: right 2 terms: st abort: F discard: F computed: T

Additional information

Used terms: st