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Proof step 5 from proof putarray-field (previous step)

is_fskey(sk), ∀ sk0, r1, i0, st0, r2. ¬ is_indexkey(sk0) → (putArray(vals0, r1, i0, st0)[r2 -- sk0]) = st0[r2 -- sk0] ⊦ (((r0 = r ∧ sk = i ') ⊃ val; putArray(vals0, r, i + 1, st)[r0 -- sk])) = st[r0 -- sk]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 3:
⊦ is_fskey(sk) → (putArray(vals, r, i, st)[r -- sk]) = st[r -- sk]
Proof step 4:
⊦ r ≠ r0 → (putArray(vals, r, j, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ is_fskey(i ')])]]