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Proof step 44 from proof addobj-addclass-sw (previous step)

is_fskey(sk) ⊦ jvmref = r, (jvmref -- sk ∈ addclass(class0, fis0, st) ↔ jvmref -- sk ∈ addclass(class0, fis0, addobj(r, class, fis, st))) ∧ (addclass(class0, fis0, st)[jvmref -- sk]) = (addclass(class0, fis0, addobj(r, class, fis, st))[jvmref -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
⊦ ¬ sk ∈ fis ∧ sk ≠ mkfs(class, void_type, `initstate' .field) ' → (r -- sk ∈ addclass(class, fis, st) ↔ r -- sk ∈ st)
Proof step 39:
¬ (¬ sk ∈ fis0 ∧ sk ≠ mkfs(class0, void_type, `initstate' .field) '), is_fskey(sk) ⊦ jvmref = r, (jvmref -- sk ∈ addclass(class0, fis0, st) ↔ jvmref -- sk ∈ addclass(class0, fis0, addobj(r, class, fis, st))) ∧ (addclass(class0, fis0, st)[jvmref -- sk]) = (addclass(class0, fis0, addobj(r, class, fis, st))[jvmref -- sk])
Proof step 40:
⊦ ¬ sk ∈ fis ∧ sk ≠ mkfs(class, void_type, `initstate' .field) ' → (r -- sk ∈ addclass(class, fis, st) ↔ r -- sk ∈ st)
Proof step 41:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 42:
⊦ ¬ sk ∈ fis ∧ sk ≠ mkfs(class, void_type, `initstate' .field) ' → (addclass(class, fis, st)[r -- sk]) = st[r -- sk]
Proof step 43:
⊦ r ≠ r0 → (addobj(r, class1, fis, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[crewrite-arg: ac-in-10 , F , [st, fis, class, sk, r] → [st, fis0, class0, sk, jvmref] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, fis0, class0, sk, jvmref)]]