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Proof step 15 from proof vals-in (previous step)

∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ val ∈ (fv + fis0) .vals ↔ (∃ sk. sk × val ∈ fv + fis0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (fv ' + fis) .vals = fv .2 ' + fis .vals
Proof step 14:
∀ val. val ∈ fis0 .vals ↔ (∃ sk. sk × val ∈ fis0) ⊦ val = fv .2 ∨ val ∈ fis0 .vals ↔ (∃ sk. sk × val = fv ∨ sk × val ∈ fis0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ fv ∈ fv0 ' + fis ↔ fv ≠ fv0 ∧ ¬ fv ∈ fis], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val ∈ val0 ' ↔ val = val0], [CSIMPSEQ : ⊦ val ∈ vals + vals0 ↔ ¬ ¬ (val ∈ vals ∨ val ∈ vals0)], [CSIMPSEQ : ⊦ (fis + fis0) + fis1 = fis + fis0 + fis1], [CSIMPSEQ : ⊦ fv + fis = fv ' + fis])]]