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Proof step 6 from proof dups-fis-keys (previous step)

¬ sk ∈ fis ⊦ dups(fis[sk, val] .keys) ↔ dups(fis .keys)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (fis + fis0) .keys = fis .keys + fis0 .keys
Proof step 3:
⊦ fv ' .keys = fv .1 '
Proof step 5:
sk ∈ fis .keys ⊦ sk ∈ fis, dups(fis .keys)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ dups(sks) → dups(sks + sks0)], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ dups(sks + sk ') ↔ sk ∈ sks ∨ dups(sks)], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ (sk × val) .1 = sk], [CSIMPSEQ : ⊦ ¬ sk ∈ fis → fis[sk, val] = fis + (sk × val) '])]]