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Proof step 5 from proof disjoint-move-left (previous step)

st0 ⊥ st1 ⊦ ∀ a0. ¬ a0 ∈ st0[a, d] ∨ ¬ a0 ∈ st1 -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ st1, a0 ∈ st0, st0 ⊥ st1 ⊦ a0 = a

Rule argument

Additional information

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