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Proof step 25 from proof ac-in-03-02 (previous step)

is_fskey(sk0), r ∈ st, r -- sk ∈ addclass(class1, (sk0 × val) ' + fis0, st), ∀ r, sk, class1, st. r -- sk ∈ addclass(class1, fis0, st) ∧ r ∈ st → r = jvmref ∨ r -- sk ∈ st ⊦ r = jvmref, r -- sk ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 17:
¬ (is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class1), is_fskey(sk0), r ∈ st, r -- sk ∈ addclass(class1, (sk0 × val) ' + fis0, st), ∀ r, sk, class1, st. r -- sk ∈ addclass(class1, fis0, st) ∧ r ∈ st → r = jvmref ∨ r -- sk ∈ st ⊦ r = jvmref, r -- sk ∈ st
Proof step 24:
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

Rule argument

[crewrite-arg: ac-no1 , F , [st, fis, fv, class] → [st, fis0, sk0 × val, class1] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, fis0, sk0 × val, class1)]]