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Proof step 14 from proof ac-dups-05 (previous step)

is_fskey((sk × val) .1) ∧ (sk × val) .1 .fs .class ≠ class1, is_fskey(sk) ⊦ addclass(class1, fis1 + (sk × val0) ', st) = addclass(class1, (sk × val) ' + fis1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
⊦ is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 13:
is_fskey(sk) ⊦ addclass(class1, fis1 + (sk × val0) ', st) = addclass(class1, fis1, st), sk .fs .class = class1

Rule argument

[emptyarg]

Additional information

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