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Proof step 15 from proof ac-in-08 (previous step)

¬ (is_fskey(fv .1) ∧ fv .1 .fs .class ≠ class1), r ∈ st, ∀ r, st, class1. r ∈ st → r ∈ addclass(class1, fis0, st) ⊦ r ∈ addclass(class1, fv ' + fis0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ is_indexkey(fv .1) → addclass(class, fv ' + fis, st) = addclass(class, fis, st)
Proof step 14:
r ∈ st, ∀ r, st, class1. r ∈ st → r ∈ addclass(class1, fis0, st) ⊦ r ∈ addclass(class1, fv ' + fis0, st), fv .1 .fs .class ≠ class1 ∧ ¬ is_indexkey(fv .1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ ¬ is_indexkey(sk) ↔ is_fskey(sk)], [CSIMPSEQ : ⊦ ¬ is_fskey(sk) ↔ is_indexkey(sk)])]]