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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ st1 ⊦ 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])

Rule argument

Additional information

Used simplifier rules: union-at-left: ⊦ (st0 ∪ st1)[a] = (a ∈ st0 ⊃ st0[a];st1[a]);%union-in: ⊦ a ∈ st0 ∪ st1 ↔ a ∈ st0 ∨ a ∈ st1;%a ≠ b ⊦ (st -- b)[a] = st[a]; (s)% ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; (ws)% ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (s)