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Proof step 15 from proof keys-put-in (previous step)

fv = sk0 × val0, sk ∈ fv ' + fis0, ∀ sk, val. sk ∈ fis0 → fis0[sk, val] .keys = fis0 .keys ⊦ (fv ' + fis0)[sk, val] .keys = sk0 ' + fis0 .keys

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ fv ' .keys = fv .1 '
Proof step 6:
⊦ (fis + fis0) .keys = fis .keys + fis0 .keys
Proof step 14:
sk ∈ fis0, ∀ sk, val. sk ∈ fis0 → fis0[sk, val] .keys = fis0 .keys ⊦ ((sk0 × val0) ' + fis0)[sk, val] .keys = sk0 ' + fis0 .keys

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : sk ≠ sk0 ⊦ sk ∈ (sk0 × val) ' + fis ↔ sk ∈ fis], [CSIMPSEQ : ⊦ ((sk × val) ' + fis)[sk, val0] = (sk × val0) ' + fis], [CSIMPSEQ : ⊦ (sk × val) .1 = sk], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ sks + sks0 = sks + sks1 ↔ sks0 = sks1], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1])]]