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Proof step 40 from proof hasSharing-dd-help (previous step)

st[r -- firstn(1 + # refs, sks0)] = refval(r1) ∧ is_referencevalue(st[r -- firstn(1 + # refs, sks0)]), # refs4 = # refs, # sks0 = 2 + # refs + # refs1, # refs1 = # refs2 + 1 + # refs0, fits(r, sks0, st), validrefs(str, r ' + refs + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, r ' + refs4 + (r1 ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs(str, (r1 ' + refs0) + r2 ' + refs2 + r0 ', st[r1 -- _type] .type, st) ⊦ r0 = jvmref, r2 = r1, r1 = r0, hasSharing(r, st), path∃(str, r2, r1, st), path∃(str, r1, r2, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
# refs1 = # refs2 + 1 + # refs0, # sks0 = 2 + # refs + # refs1, # refs4 = # refs, st[r -- firstn(1 + # refs, sks0)] = refval(r1), validrefs(str, (r1 ' + refs0) + r2 ' + refs2 + r0 ', st[r1 -- _type] .type, st), validrefs(str, r ' + refs4 + (r1 ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs(str, r ' + refs + r1 ' + refs1 + r0 ', st[r -- _type] .type, st), fits(r, sks0, st) ⊦ r1 = r0, r2 = r1, r0 = jvmref, path∃(str, r1, r2, st), path∃(str, r2, r1, st), hasSharing(r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_referencevalue(refval(r))])]]