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Proof step 18 from proof ac-in-11 (previous step)

r -- sk ∈ addclass(class, (sk1 × val) ' + fis0, st), is_indexkey(sk), is_fskey(sk1), r ∈ addclass(class, (sk1 × val) ' + fis0, st), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ 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((sk1 × val) .1) ∧ (sk1 × val) .1 .fs .class ≠ class), r -- sk ∈ addclass(class, (sk1 × val) ' + fis0, st), is_indexkey(sk), is_fskey(sk1), r ∈ addclass(class, (sk1 × val) ' + fis0, st), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ st
Proof step 17:
is_fskey((sk1 × val) .1) ∧ (sk1 × val) .1 .fs .class ≠ class, r -- sk ∈ addclass(class, fis0, st), is_indexkey(sk), is_fskey(sk1), r ∈ addclass(class, fis0, st), ∀ sk0, r0, class0, st0. ¬ is_fskey(sk0) → (r0 -- sk0 ∈ addclass(class0, fis0, st0) → r0 -- sk0 ∈ st0) ⊦ r -- sk ∈ st

Rule argument

[crewrite-arg: ac-no1 , F , [st, fis, fv, class] → [st, fis0, sk1 × val, class] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, fis0, sk1 × val, class)]]