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Proof step 43 from proof put-switch-primval-index-index (previous step)

¬ ( ¬ is_referencevalue(st[r1 -- sk]) ∧ fits(r0, sks, st) ∧ ¬ junkref(str, r1, st) ∧ ¬ junkref(str, r0, st) ∧ is_referencevalue(st[r0 -- sks]) ∧ sks ≠ []), st[r -- sks1] = refval(r1), st[r0 -- sks] = refval(r2), fits(r0, sks + sk0 ', st), fits(r, sks1 + sk ', st), is_indexkey(sk0), is_indexkey(sk), r1 ≠ r2 ∨ sk ≠ sk0 ⊦ (st[r1 -- sk, val][(st[r1 -- sk, val][r0 -- sks]) .refval -- sk0, val0]) = (st[r2 -- sk0, val0][(st[r2 -- sk0, val0][r -- sks1]) .refval -- sk, val]), is_referencevalue(val), is_referencevalue(val0), is_referencevalue(st[r -- (sks1 + sk ')]), is_referencevalue(st[r0 -- (sks + sk0 ')]), junkref(str, r, st), junkref(str, r0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ ¬ is_referencevalue(st[r -- sks]) ∧ ¬ junkref(str, r, st) → ¬ fits(r, sks + sk ', st)
Proof step 42:
st[r0 -- sks] = refval(r2), st[r -- sks1] = refval(r1), is_indexkey(sk), is_indexkey(sk0), fits(r, sks1 + sk ', st), fits(r0, sks + sk0 ', st), r1 ≠ r2 ∨ sk ≠ sk0 ⊦ (st[r1 -- sk, val][(st[r1 -- sk, val][r0 -- sks]) .refval -- sk0, val0]) = (st[r2 -- sk0, val0][(st[r2 -- sk0, val0][r -- sks1]) .refval -- sk, val]), junkref(str, r0, st), junkref(str, r, st), is_referencevalue(st[r0 -- (sks + sk0 ')]), is_referencevalue(st[r -- (sks1 + sk ')]), is_referencevalue(val0), is_referencevalue(val), sks ≠ [] ∧ ¬ junkref(str, r1, st) ∧ fits(r0, sks, st) ∧ ¬ is_referencevalue(st[r1 -- sk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : javavalue, , test-19, ⊦ is_referencevalue(refval(r))], [CSIMPNAMEDSEQ : list, storekeys, a, ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1])]]