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Proof step 7 from proof union-move-left (previous step)

a ∈ st1 ⊦ a0 = a, a ∈ st0, (a0 ∈ st0[a, st1[a]] ∪ (st1 -- a) ↔ a0 ∈ st0 ∪ st1) ∧ (a0 ∈ st0[a, st1[a]] ⊃ st0[a, st1[a]][a0];st1[a0]) = (a0 ∈ st0 ⊃ st0[a0];st1[a0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ st1 ⊦ a0 ∈ st0[a, st1[a]] ∪ (st1 -- a) ↔ a0 ∈ st0 ∪ st1, a0 = a, a ∈ st0

Rule argument

Additional information

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