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Proof step 13 from proof vals-in (previous step)

fv = sk0 × val0, ∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ val = val0 ∨ val ∈ fis0 .vals ↔ (∃ sk. sk × val = fv ∨ sk × val ∈ fis0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ val = val0 ∨ val ∈ fis0 .vals ↔ (∃ sk. sk × val = sk0 × val0 ∨ sk × val ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]