[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 43 from proof subset-is-union (previous step)

st1[rk, st0[rk]] = st ∪ st0, st0 .keys ∩ st .keys = ∅, rk ∈ 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 = rk, st1[rk0] = ((st -- rk) ∪ st0)[rk0], ∃ 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 32:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 42:
∀ rk0. (rk0 ∈ st1[rk, st0[rk]] ↔ rk0 ∈ st ∪ st0) ∧ (st1[rk, st0[rk]][rk0]) = (st ∪ st0)[rk0], st0 .keys ∩ st .keys = ∅, rk ∈ 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 = rk, st1[rk0] = ((st -- rk) ∪ st0)[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 ∪ st0, st1[rk, st0[rk]]] ,F ,store ,javastore ,Extension ]

Additional information

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