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Proof step 24 from proof del-put-switch (previous step)

⊦ r = r0, (rk ∈ st[r -- sk, val] -- r0 ↔ rk = r -- sk ∨ rk ∈ st -- r0) ∧ (st[r -- sk, val] -- r0)[rk] = ((rk = r -- sk ⊃ val; (st -- r0)[rk]))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
⊦ rk ∈ st[r -- sk, val] -- r0 ↔ rk = r -- sk ∨ rk ∈ st -- r0, r = r0
Proof step 23:
⊦ (st[r -- sk, val] -- r0)[rk] = ((rk = r -- sk ⊃ val; (st -- r0)[rk])), r = r0

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]