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Proof step 44 from proof hasSharing-dd-help (previous step)

refs3 = r1 + refs0, # refs4 = # refs, # sks0 = 2 + # refs + # refs1, # refs1 = # refs3 + # refs2, st[r -- firstn(1 + # refs, sks0)] .refval = r1, validrefs(str, (r ' + refs4 + refs3) + r2 ' + refs2 + r0 ', st[r -- _type] .type, st), fits(r, sks0, st), validrefs(str, r ' + refs + r1 ' + refs1 + r0 ', st[r -- _type] .type, st) ⊦ r0 = jvmref, refs4 + refs3 = [], refs3 = [], 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 43:
# 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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ refs + r ' + refs0 ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]