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Proof step 115 from proof putarray-addarray-01 (previous step)

0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0 ⊦ putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st)) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[r -- _length, intval(i0 + i + n→i(# vals))])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 114:
0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0 ⊦ ∀ rk. ( rk ∈ putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st)) ↔ rk ∈ (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st) [r -- _length, intval(i0 + i + n→i(# vals)) ] )) ∧ (putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))[rk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[r -- _length, intval(i0 + i + n→i(# vals))] [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 + mklist(initial_value(ty), i→n(i0)), st) [r -- _length, intval(i0 + i + n→i(# vals)) ], putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray (r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st) [r -- _length, intval(i0 + i + n→i(# vals)) ], putArray (vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st)))]]