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Proof step 14 from proof put-switch-primval-index-index (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
st[r -- sks1] = refval(r1), is_indexkey(sk), is_indexkey(sk0), fits(r, sks1 + sk ', st), fits(r2, sk0 ', st), r2 ≠ r1 ∨ sk0 ≠ sk, r1 ≠ r2 ∨ sk ≠ sk0 ⊦ sks1 = [], (st[r2 -- sk0, val0][r1 -- sk, val]) = (st[r2 -- sk0, val0][(st[r2 -- sk0, val0][r -- sks1]) .refval -- sk, val]), is_referencevalue(st[r -- (sks1 + sk ')]), is_referencevalue(val0), is_referencevalue(val), junkref(str, r2, st), junkref(str, r, st), is_referencevalue(st[r2 -- sk0]), fits(r, sks1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]