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Proof step 39 from proof sharing-twopaths (conclusion)

⊦ sks ≠ [] ∧ sks0 ≠ [] ∧ st[r -- sks] .refval ≠ jvmref ∧ st[r -- sks0] .refval ≠ jvmref ∧ r ≠ jvmref ∧ path∃(str, st[r -- sks] .refval, st[r -- sks0] .refval, st) ∧ sks ≠ sks0 ∧ fits(r, sks, st) ∧ fits(r, sks0, st) ∧ ¬ sks ⊑ sks0 → hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
fits(r, sks0, st), fits(r, sks, st), path∃(str, st[r -- sks] .refval, st[r -- sks0] .refval, st) ⊦ sks = [], sks0 = [], st[r -- sks] .refval = jvmref, st[r -- sks0] .refval = jvmref, r = jvmref, sks = sks0, sks ⊑ sks0, hasSharing(r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]