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Proof step 5 from proof update-keyin-01 (previous step)

r -- sk ∈ update(r0, fis0, st) ↔ r = r0 ∧ sk ∈ fis0, r = r0 ∧ sk = sk0 ∨ r = r0 ∧ sk ∈ fis0, ∀ r, sk, r0, st. r -- sk ∈ update(r0, fis0, st) ↔ r -- sk ∈ st ∨ r = r0 ∧ sk ∈ fis0 ⊦ r -- sk ∈ st, r = r0 ∧ (sk = sk0 ∨ sk ∈ fis0)

User applied rule case distinction

And obtained the following new goals

Proof step 4:
⊦ ¬ sk ∈ fis → (r0 -- sk ∈ update(r, fis, st) ↔ r0 -- sk ∈ st)

Rule argument

[CASEDARG : pos: left 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]