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Proof step 75 from proof hasSharing-dd-help (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ m + n = n + m])]]