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Proof step 6 from proof Data-notin-other-3 (previous step)

a0 ∈ st[a, d1][b, d2] ⊦ st[a, d1][b, d2][a0] = d1, st[a, d1][b, d2][a0] = d2, a0 = c, a0 = b, d = st[a, d1][b, d2][a0], ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][b, d2][a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ∈ st[a, d1] ⊦ st[a, d1][a0] = d2, st[a, d1][a0] = d1, a0 = b, a0 = c, d = st[a, d1][a0], ∃ a1. a1 ∈ st ∧ st[a1] = st[a, d1][a0]

Rule argument

Additional information

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