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Proof step 4 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, a ∈ st[a, d1] ∧ st[a, d1][a] = d0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ st ⊦ d1 = st[a0], a ∈ st, ∃ a1. a1 ∈ st[a, d1] ∧ st[a, d1][a1] = st[a0]

Rule argument

Additional information

Used simplifier rules: ⊦ st[a, d][a] = d; (s)% ⊦ a ∈ st[a, d]; (s)