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Proof step 12 from proof vals-in (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 11:
val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0), ∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ val = val0 ∨ val ∈ fis0 .vals ↔ (∃ sk. sk × val = sk0 × val0 ∨ sk × val ∈ fis0)

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (val)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]