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Proof step 29 from proof hassharing-validrefs-dups (previous step)

path∃(str, r, r, st), validrefs(str, r ' + refs1 + r ' + refs, st[r -- _type] .type, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)

User applied rule cut formula

And obtained the following new goals

Proof step 22:
validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), path∃(str, r, r, st), validrefs(str, r ' + refs1 + r ' + refs, st[r -- _type] .type, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)
Proof step 28:
path∃(str, r, r, st), validrefs(str, r ' + refs1 + r ' + refs, st[r -- _type] .type, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), hasSharing(r, st)

Rule argument

[fma-arg: validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st)]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (validrefs (str, r ' + refs1 + r ', st[r -- _type] .type, st))]]