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Proof step 99 from proof hassharing-validrefs-dups (previous step)

validrefs(str, r ' + refs0, st[r -- _type] .type, st), path∃(str, r, r, st), dups(refs0), ∀ st, str, refs, r0. # (r0 ' + refs) < # (r ' + refs0) → (validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ refs0 = [], r ∈ refs0, hasSharing(r, st)

User applied rule cut formula

And obtained the following new goals

Proof step 94:
validrefs(str, refs0, st[refs0 .first -- _type] .type, st), validrefs(str, r ' + refs0, st[r -- _type] .type, st), path∃(str, r, r, st), dups(refs0), ∀ st, str, refs, r0. # (r0 ' + refs) < # (r ' + refs0) → (validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ refs0 = [], r ∈ refs0, hasSharing(r, st)
Proof step 98:
validrefs(str, r ' + refs0, st[r -- _type] .type, st), path∃(str, r, r, st), dups(refs0), ∀ st, str, refs, r0. # (r0 ' + refs) < # (r ' + refs0) → (validrefs(str, r0 ' + refs, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs ∧ ¬ dups(refs)) ⊦ validrefs(str, refs0, st[refs0 .first -- _type] .type, st), refs0 = [], r ∈ refs0, hasSharing(r, st)

Rule argument

[fma-arg: validrefs(str, refs0, st[refs0 .first -- _type] .type, st)]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (validrefs (str, refs0, st[refs0 .first -- _type] .type, st))]]