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Proof step 10 from proof update-in-03 (previous step)

fv = sk × val, ∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r ∈ (update(r0, fis0, st)[r0 -- sk, val]) ↔ r ∈ st ∨ r = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r ∈ (update(r0, fis0, st)[r0 -- sk, val]) ↔ r ∈ st ∨ r = r0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]