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Proof step 22 from proof ac-lookup-05 (previous step)

is_indexkey(sk), is_fskey(sk0), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 15:
¬ (is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class), is_indexkey(sk), is_fskey(sk0), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(class, (sk0 × val) ' + fis0, st)[r -- sk]) = st[r -- sk]
Proof step 21:
is_fskey((sk0 × val) .1) ∧ (sk0 × val) .1 .fs .class ≠ class, is_indexkey(sk), is_fskey(sk0), ∀ sk0, class0, st0, r0. ¬ is_fskey(sk0) → (addclass(class0, fis0, st0)[r0 -- sk0]) = st0[r0 -- sk0] ⊦ (addclass(class, fis0, st)[r -- sk]) = st[r -- sk]

Rule argument

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

Additional information

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