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Proof step 35 from proof del-exists (previous step)

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

User applied rule exists right

And obtained the following new goals

Proof step 36:
st0 = st[a, st0[a]], a ∈ st0, ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ a = a0, a ∈ st, ¬ a ∈ st[a0, d] ∧ st0[a0, d] = st[a0, d][a, st0[a]], ∃ st1. ¬ a ∈ st1 ∧ st0[a0, d] = st1[a, st0[a]]

Rule argument

pos: right 3 terms: st[a0, d] abort: F discard: F computed: F

Additional information

Used terms: st[a0, d]