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

val = val0 ∨ val ∈ fis0 .vals, 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

User applied rule case distinction

And obtained the following new goals

Proof step 4:
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
Proof step 6:
val ∈ fis0 .vals, 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

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]