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Proof step 42 from proof hasSharing-dd-help (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
# refs4 = # refs, # sks0 = 2 + # refs + # refs1, # refs1 = # refs2 + 1 + # refs0, fits(r, sks0, st), validrefs(str, r ' + refs + st[r -- firstn(1 + # refs, sks0)] .refval ' + refs1 + r0 ', st[r -- _type] .type, st), validrefs(str, r ' + refs4 + (st[r -- firstn(1 + # refs, sks0)] .refval ' + refs0) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), validrefs (str, (st[r -- firstn(1 + # refs, sks0)] .refval ' + refs0) + r2 ' + refs2 + r0 ', st[st[r -- firstn(1 + # refs, sks0)] .refval -- _type] .type, st) ⊦ r0 = jvmref, r2 = st[r -- firstn(1 + # refs, sks0)] .refval, st[r -- firstn(1 + # refs, sks0)] .refval = r0, hasSharing(r, st), path∃(str, r2, st[r -- firstn(1 + # refs, sks0)] .refval, st), path∃(str, st[r -- firstn(1 + # refs, sks0)] .refval, r2, st)

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])]]