[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 23 from proof refs-equal (previous step)

# st .refs = # st0 .refs, rk .ref ∈ st0 .refs, rk .ref ∈ st .refs, r ∈ st0, rk ∈ st -- rk .ref, rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r = rk .ref, st0 = ∅, st = ∅, r ∈ st, rk ∈ st0 -- rk .ref ∧ (st -- rk .ref)[rk] = (st0 -- rk .ref)[rk]

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r0, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk ∈ st -- r0 ↔ r ≠ r0 ∧ r -- sk ∈ st], [CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r0, sk)]]