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Proof step 19 from proof switch-array (previous step)

# vals = m + 1, ∀ st, vals0, ty0, ty, r0, r, vals1. # vals1 < m + 1 → (true → r = r0 ∨ addarray(r, ty, vals1, addarray(r0, ty0, vals0, st)) = addarray(r0, ty0, vals0, addarray(r, ty, vals1, st))) ⊦ r0 = r, (addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last])[r -- _length, intval(n→i(m) + 1)]) = addarray(r0, ty0, vals0, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last][r -- _length, intval(n→i(m) + 1)]), vals = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ r ≠ r0 → (addarray(r, ty, vals, st)[r0 -- sk, val]) = addarray(r, ty, vals, st[r0 -- sk, val])

Rule argument

[crewrite-arg: shift-1init , F , [val, sk, r0, st, vals, ty, r] → [intval(n→i(m) + 1), _length, r, addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last], vals0, ty0, r0] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (intval(n→i(m) + 1), _length, r, addarray(r, ty, vals .butlast, st) [r -- n→i(m) ', vals .last ], vals0, ty0, r0)]]