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Proof step 53 from proof subset-is-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ rk ∈ st .keys → (st -- rk) .keys = st .keys -- rk
Proof step 14:
¬ rk ∈ st .keys, st1[rk, st0[rk]] = st ∪ st0, st0 .keys ∩ st .keys = ∅, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ st1 = ((st -- rk) ∪ st0)[rk, st1[rk]] ∧ (st -- rk) .keys ∩ (st0 .keys ++ rk) = ∅, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅
Proof step 52:
rk ∈ st .keys, st1[rk, st0[rk]] = st ∪ st0, st0 .keys ∩ st .keys = ∅, st0 ⊆ st1[rk, st0[rk]], rk ∈ st1, st0 -- rk ⊆ st1 -- rk, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ st1 = ((st -- rk) ∪ st0)[rk, st1[rk]] ∧ (st .keys -- rk) ∩ (st0 .keys ++ rk) = ∅, ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[speclemmaarg: ,rk ∈ st .keys → (st -- rk) .keys = st .keys -- rk ,[rk, st] → [rk, st] ,F ,ostore ,javastore ,del-01 ]

Additional information

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