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Proof step 16 from proof copy-succ (previous step)

0 ≤ j ⊦ (((i = i0 + j ∧ r = r0) ⊃ st[r1 -- (i1 + j) ']; arraycopy(r1, i1, r0, i0, j, st)[r -- i '])) = (arraycopy(r1, i1, r0, i0, j, st[r0 -- (i0 + j) ', st[r1 -- (i1 + j) ']])[r -- i ']), r1 = r0

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 5:
⊦ (arraycopy(r, i, r0, i0, j, st)[r0 -- (i0 + j) ']) = st[r0 -- (i0 + j) ']
Proof step 15:
i ≠ i0 + j ∨ r ≠ r0, 0 ≤ j ⊦ (arraycopy(r1, i1, r0, i0, j, st)[r -- i ']) = (arraycopy(r1, i1, r0, i0, j, st[r0 -- (i0 + j) ', st[r1 -- (i1 + j) ']])[r -- i ']), r1 = r0

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0])]]