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Proof step 18 from proof ac-in-02 (previous step)

r -- i ' ∈ addclass(class1, (sk × val) ' + fis0, st), is_fskey(sk), r ∈ addclass(class1, (sk × val) ' + fis0, st), ∀ r0, i0, class, st0. r0 -- i0 ' ∈ addclass(class, fis0, st0) → r0 -- i0 ' ∈ st0 ⊦ r -- i ' ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 11:
¬ (is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1), r -- i ' ∈ addclass(class1, (sk × val) ' + fis0, st), is_fskey(sk), r ∈ addclass(class1, (sk × val) ' + fis0, st), ∀ r0, i0, class, st0. r0 -- i0 ' ∈ addclass(class, fis0, st0) → r0 -- i0 ' ∈ st0 ⊦ r -- i ' ∈ st
Proof step 17:
is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1, r -- i ' ∈ addclass(class1, fis0, st), is_fskey(sk), r ∈ addclass(class1, fis0, st), ∀ r0, i0, class, st0. r0 -- i0 ' ∈ addclass(class, fis0, st0) → r0 -- i0 ' ∈ st0 ⊦ r -- i ' ∈ 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)]]