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Proof step 75 from proof subset-is-union (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st[rk, val] ⊆ st0 ↔ st -- rk ⊆ st0 -- rk ∧ rk × val ∈ st0
Proof step 74:
st0 -- rk ⊆ st1 -- rk ∧ rk × val ∈ st1, ∀ st1. st0 ⊆ st1 → (∃ st2. st1 = st2 ∪ st0 ∧ st0 .keys ∩ st2 .keys = ∅) ⊦ ∃ st. st1 = (st ∪ st0)[rk, val] ∧ st .keys ∩ (st0 .keys ++ rk) = ∅

Rule argument

[speclemmaarg: ,st[rk, val] ⊆ st0 ↔ st -- rk ⊆ st0 -- rk ∧ rk × val ∈ st0 ,[st0, val, rk, st] → [st1, val, rk, st0] ,F ,store ,javastore ,subset-put-02 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1, val, rk, st0)]]