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Proof step 6 from proof del-exists (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 7:
a0 ∈ st0 → (∃ st1. ¬ a0 ∈ st1 ∧ st0 = st1[a0, st0[a0]]), ∀ a. a ∈ st0 → (∃ st1. ¬ a ∈ st1 ∧ st0 = st1[a, st0[a]]) ⊦ ∃ st. ¬ a0 ∈ st ∧ st0[a0, d] = st[a0, d]

Rule argument

pos: left 1 terms: a0 abort: F discard: F computed: T

Additional information

Used terms: a0