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Proof step 6 from proof add-at-05 (previous step)

is_fskey(sk), 0 ≤ i1, ∀ r0, st, val, r, sk, i0. i→n(i0) < i→n(i1 + 1) → (is_fskey(sk) → (addarray(r, val, i0, st)[r0 -- sk]) = st[r0 -- sk]) ⊦ (((r0 = r ∧ sk = i1 ') ⊃ val; addarray(r, val, i1, st)[r0 -- sk])) = st[r0 -- sk]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
r0 ≠ r ∨ sk ≠ i1 ', is_fskey(sk), 0 ≤ i1, ∀ r0, st, val, r, sk, i0. i→n(i0) < i→n(i1 + 1) → (is_fskey(sk) → (addarray(r, val, i0, st)[r0 -- sk]) = st[r0 -- sk]) ⊦ (addarray(r, val, i1, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

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