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Proof step 5 from proof get-eqex-03 (previous step)

j0 < j, st -- r0 = st0 -- r0, eqex(st0, st, r0), 0 ≤ j0, 0 < j ⊦ st0[r -- (i + j0) '] = st[r -- (i + j0) '], r = r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk])
Proof step 4:
j0 < j, ∀ rk. (rk ∈ st -- r0 ↔ rk ∈ st0 -- r0) ∧ (st -- r0)[rk] = (st0 -- r0)[rk], eqex(st0, st, r0), 0 ≤ j0, 0 < j ⊦ st0[r -- (i + j0) '] = st[r -- (i + j0) '], r = r0

Rule argument

[speclemmaarg: ,st1 = st2 ↔ (∀ rk. (rk ∈ st1 ↔ rk ∈ st2) ∧ st1[rk] = st2[rk]) ,[st2, st1] → [st0 -- r0, st -- r0] ,F ,store ,javastore ,Extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0 -- r0, st -- r0)]]