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Proof step 5 from proof add-at-05 (previous step)

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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 → (addarray(r, val, i, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 4:
0 ≤ i1, is_fskey(sk), ∀ 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]) ⊦ sk = i1 ', (addarray(r, val, i1, st)[r -- sk]) = st[r -- sk]

Rule argument

[emptyarg]

Additional information

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