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Proof step 35 from proof copy-succ (previous step)

0 ≤ j ⊦ r = r0, (arraycopy(r, i, r0, i0, j, st)[r0 -- (i0 + j) ', st[r -- (i + j) ']]) = arraycopy(r, i, r0, i0, j, st[r0 -- (i0 + j) ', st[r -- (i + j) ']])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 34:
0 ≤ j ⊦ r = r0, ∀ rk. ( rk ∈ (arraycopy(r, i, r0, i0, j, st)[r0 -- (i0 + j) ', st[r -- (i + j) ']]) ↔ rk ∈ arraycopy(r, i, r0, i0, j, st[r0 -- (i0 + j) ', st[r -- (i + j) ']])) ∧ (arraycopy(r, i, r0, i0, j, st)[r0 -- (i0 + j) ', st[r -- (i + j) ']][rk]) = (arraycopy(r, i, r0, i0, j, st[r0 -- (i0 + j) ', st[r -- (i + j) ']])[rk])

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [arraycopy (r, i, r0, i0, j, st[r0 -- (i0 + j) ', st[r -- (i + j) ']]), arraycopy(r, i, r0, i0, j, st) [r0 -- (i0 + j) ', st[r -- (i + j) '] ]] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (arraycopy (r, i, r0, i0, j, st[r0 -- (i0 + j) ', st[r -- (i + j) ']]), arraycopy(r, i, r0, i0, j, st) [r0 -- (i0 + j) ', st[r -- (i + j) '] ])]]