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Proof step 6 from proof infis-notin-01 (previous step)

∀ st, r, sk, val. (fis0 ∈ st[r -- sk, val] ↔ fis0 ∈ st) ∨ sk ∈ fis0 ⊦ (fv + fis0 ∈ st[r -- sk, val] ↔ fv + fis0 ∈ st) ∨ sk ∈ fv + fis0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ st, r, sk, val. (fis0 ∈ st[r -- sk, val] ↔ fis0 ∈ st) ∨ sk ∈ fis0 ⊦ fv ' + fis0 ∈ st[r -- sk, val] ↔ fv ' + fis0 ∈ st, sk ∈ fv ' + fis0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ fv + fis = fv ' + fis])]]