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Proof step 23 from proof addclass-switch (previous step)

is_fskey((sk × val0) .1) ∧ (sk × val0) .1 .fs .class ≠ class, is_fskey(sk), ∀ class, st, r, i, val. addclass(class, fis0, st[r -- i ', val]) = (addclass(class, fis0, st)[r -- i ', val]) ⊦ addclass(class, fis0, st[r -- i ', val]) = (addclass(class, (sk × val0) ' + fis0, st)[r -- i ', val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 22:
is_fskey(sk), ∀ class, st, r, i, val. addclass(class, fis0, st[r -- i ', val]) = (addclass(class, fis0, st)[r -- i ', val]) ⊦ addclass(class, fis0, st[r -- i ', val]) = (addclass(class, fis0, st)[r -- i ', val]), sk .fs .class = class

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sk × val) .1 = sk], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1])]]