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Proof step 13 from proof subset-is-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ rk ∈ st .keys ↔ rk ∈ st
Proof step 12:
st0 .keys ∩ st .keys = ∅, st1[rk, st0[rk]] = st ∪ st0, st0 -- rk ⊆ st1 -- rk, rk ∈ st1, st0 ⊆ st1[rk, st0[rk]], ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ rk ∈ st, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅, st1 = ((st -- rk) ∪ st0)[rk, st1[rk]] ∧ (st -- rk) .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[speclemmaarg: ,rk ∈ st .keys ↔ rk ∈ st ,[st, rk] → [st, rk] ,F ,ostore ,javastore ,Keys ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, rk)]]