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Proof step 52 from proof addobj-addclass-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
⊦ r ≠ r0 → (r -- sk ∈ addobj(r0, class1, fis, st) ↔ r -- sk ∈ st)
Proof step 51:
is_indexkey(sk) ⊦ jvmref = r, (addclass(class0, fis0, st)[jvmref -- sk]) = (addclass(class0, fis0, addobj(r, class, fis, st))[jvmref -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_fskey(sk) ↔ is_indexkey(sk)])]]