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Proof step 48 from proof subset-is-union (previous step)

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. 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 15:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 47:
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 = ∅) ⊦ ∀ rk0. (rk0 ∈ st1 ↔ rk0 ∈ ((st -- rk) ∪ st0)[rk, st1[rk]]) ∧ st1[rk0] = (((st -- rk) ∪ st0)[rk, st1[rk]][rk0]), ∃ st. st1 = (st ∪ st0)[rk, st1[rk]] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [((st -- rk) ∪ st0)[rk, st1[rk]], st1] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (((st -- rk) ∪ st0)[rk, st1[rk]], st1)]]