[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 9 from proof hasKeys-all (previous step)

∀ sk0. sk0 = sk ∨ sk0 ∈ sks0 → r -- sk0 ∈ st, hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st), ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st) ⊦ r -- sk ∈ st ∧ hasKeys(r, sks0, st)

User applied rule all left

And obtained the following new goals

Proof step 8:
sk = sk ∨ sk ∈ sks0 → r -- sk ∈ st, ∀ sk0. sk0 ∈ sks0 → r -- sk0 ∈ st, hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st), ∀ r, st. hasKeys(r, sks0, st) ↔ (∀ sk. sk ∈ sks0 → r -- sk ∈ st) ⊦ r -- sk ∈ st ∧ hasKeys(r, sks0, st)

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (sk)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk)]]