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Proof step 2 from proof Data-in-other (previous step)

a0 ∈ st ∧ st[a0] = d0 ∨ d0 = d1 ⊦ a ∈ st, ∃ a0. a0 ∈ st[a, d1] ∧ st[a, d1][a0] = d0

User applied rule exists right

And obtained the following new goals

Proof step 3:
a0 ∈ st ∧ st[a0] = d0 ∨ d0 = d1 ⊦ a ∈ st, a ∈ st[a, d1] ∧ st[a, d1][a] = d0, ∃ a0. a0 ∈ st[a, d1] ∧ st[a, d1][a0] = d0

Rule argument

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

Additional information

Used terms: a