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Proof step 23 from proof del-put-switch (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
⊦ r ≠ r0 → (st -- r)[r0 -- sk0] = st[r0 -- sk0]
Proof step 22:
rk ≠ r -- sk ⊦ (st[r -- sk, val] -- r0)[rk] = (st -- r0)[rk], r = r0

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))])]]