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Proof step 30 from proof copy-succ (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 16:
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
Proof step 29:
0 ≤ j ⊦ r -- i ' ∈ (arraycopy(r1, i1, r0, i0, j, st)[r0 -- (i0 + j) ', st[r1 -- (i1 + j) ']]) ↔ r -- i ' ∈ arraycopy(r1, i1, r0, i0, j, st[r0 -- (i0 + j) ', st[r1 -- (i1 + j) ']]), r1 = r0

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]