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Proof step 226 from proof putarray-addarray-01 (previous step)

0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ putArray(vals, r, i, addarray(r, ty, j, st)) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- _length, intval(j)])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 119:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 225:
0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ ∀ rk. (rk ∈ putArray(vals, r, i, addarray(r, ty, j, st)) ↔ rk ∈ (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- _length, intval(j)])) ∧ (putArray(vals, r, i, addarray(r, ty, j, st))[rk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- _length, intval(j)][rk])

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st) [r -- _length, intval(j) ], putArray(vals, r, i, addarray(r, ty, j, st))] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray (r, ty, mklist(initial_value(ty), i→n(i)) + vals, st) [r -- _length, intval(j) ], putArray(vals, r, i, addarray(r, ty, j, st)))]]