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Proof step 11 from proof hasSharing-rec (previous step)

hasSharing(r, st) ∨ r0 ∈ refs ∧ hasSharing(r0, st) ⊦ (r0 = r ∨ r0 ∈ refs) ∧ hasSharing(r0, st), ∃ r1. (r1 = r ∨ r1 ∈ refs) ∧ hasSharing(r1, st)

User applied rule case distinction

And obtained the following new goals

Proof step 10:
hasSharing(r, st) ⊦ (r0 = r ∨ r0 ∈ refs) ∧ hasSharing(r0, st), ∃ r1. (r1 = r ∨ r1 ∈ refs) ∧ hasSharing(r1, st)

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]