[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 32 from proof addobj-addobj-switch (previous step)

⊦ r = r0, r1 = r, (r1 -- sk ∈ addobj(r0, class0, fis0, st) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r0, class0, fis0, st)[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 17:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 29:
¬ r1 ≠ r0 ⊦ r = r0, r1 = r, (r1 -- sk ∈ addobj(r0, class0, fis0, st) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r0, class0, fis0, st)[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])
Proof step 30:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 31:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[crewrite-arg: add-in-04 , F , [st, fis, class1, r0, sk, r] → [addobj(r, class, fis, st), fis0, class0, r0, sk, r1] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addobj(r, class, fis, st), fis0, class0, r0, sk, r1)]]