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Proof step 21 from proof ac-in-09 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 12:
¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1), is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, (sk × val) ' + fis0, st) ↔ r ∈ st
Proof step 20:
is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1, is_fskey(sk), ∀ r, class1, st. r = jvmref ∨ (r ∈ addclass(class1, fis0, st) ↔ r ∈ st) ⊦ r = jvmref, r ∈ addclass(class1, fis0, st) ↔ r ∈ st

Rule argument

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

Additional information

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