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Proof step 58 from proof subset-is-union (previous step)

rk0 ∈ st0, rk ∈ st1, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅), ∀ rk1. rk1 ∈ st0 -- rk → rk1 ≠ rk ∧ rk1 ∈ st1 ∧ st0[rk1] = st1[rk1] ⊦ rk0 ∈ st1, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

[EXRARG : pos: left 4, [QUANTTERMLIST : (rk0)]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ st -- rk0 ↔ rk ≠ rk0 ∧ rk ∈ st])], [EXTRATERMS : (rk0)]]