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Proof step 7 from proof keys-in (previous step)

∀ sk. sk ∈ fis0 .keys ↔ sk ∈ fis0 ⊦ sk ∈ (fv + fis0) .keys ↔ sk ∈ fv + fis0

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 6:
∀ sk. sk ∈ fis0 .keys ↔ sk ∈ fis0 ⊦ sk = fv .1 ∨ sk ∈ fis0 .keys ↔ sk ∈ fv ' ∨ sk ∈ fis0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ∈ fis + fis0 ↔ sk ∈ fis ∨ sk ∈ fis0], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk ∈ sk0 ' ↔ sk = sk0], [CSIMPSEQ : ⊦ sk ∈ sks + sks0 ↔ ¬ ¬ (sk ∈ sks ∨ sk ∈ sks0)], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ fv + fis = fv ' + fis])]]