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Proof step 5 from proof update-in-03 (previous step)

r = r0 ∨ r ∈ update(r0, fis0, st), ∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r ∈ st ∨ r = r0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
r ∈ update(r0, fis0, st), ∀ r, r0, st. r ∈ update(r0, fis0, st) ↔ r ∈ st ∨ r = r0 ∧ fis0 ≠ [] ⊦ r = r0, r ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]