[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 10 from proof is_array-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r ∈ st1, type∃(str, st1[r -- _type] .type), is_array(r, st1), ¬ type∃(str, (st1 ∪ st)[r -- _type] .type) ∨ (st1 ∪ st)[r -- _type] .type .jtelemtype = void_type ∨ (st1 ∪ st)[r -- _type] .type .jtelemtype = abstract_type ⊦ st1[r -- _type] .type .jtelemtype = void_type, st1[r -- _type] .type .jtelemtype = abstract_type

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk]
Proof step 9:
¬ ¬ r -- _type ∈ st, st1 .refs ∩ st .refs = ∅, r ∈ st1, type∃(str, st1[r -- _type] .type), is_array(r, st1), ¬ type∃(str, (st1 ∪ st)[r -- _type] .type) ∨ (st1 ∪ st)[r -- _type] .type .jtelemtype = void_type ∨ (st1 ∪ st)[r -- _type] .type .jtelemtype = abstract_type ⊦ st1[r -- _type] .type .jtelemtype = void_type, st1[r -- _type] .type .jtelemtype = abstract_type

Rule argument

[speclemmaarg: ,¬ rk ∈ st0 → (st ∪ st0)[rk] = st[rk] ,[rk, st0, st] → [r -- _type, st, st1] ,F ,storefunsx , ,union-at-left ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r -- _type, st, st1)]]