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Proof step 4 from proof vals-in (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
val0 ∈ fis0 .vals ↔ (∃ sk. sk × val0 ∈ fis0), ∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ ∃ sk. sk × val0 = sk0 × val0 ∨ sk × val0 ∈ fis0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]