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Proof step 24 from proof ac-in-03-02 (previous step)

is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class1, is_fskey(sk0), r ∈ st, r -- sk ∈ addclass(class1, fis0, st), ∀ r, sk, class1, st. r -- sk ∈ addclass(class1, fis0, st) ∧ r ∈ st → r = jvmref ∨ r -- sk ∈ st ⊦ r = jvmref, r -- sk ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
⊦ r ∈ st → r ∈ addclass(class1, fis, st)
Proof step 23:
r -- sk ∈ addclass(class1, fis0, st), r ∈ st, is_fskey(sk0), ∀ r, sk, class1, st. r -- sk ∈ addclass(class1, fis0, st) ∧ r ∈ st → r = jvmref ∨ r -- sk ∈ st ⊦ sk0 .fs .class = class1, r = jvmref, r -- sk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ (sk × val) .1 = sk])]]