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Proof step 18 from proof pathex-levelsubset-01 (previous step)

validref(str, r, ty, st), path∃(str0, r, r0, st), level⊆(str, str0) ⊦ path∃(str, r, r0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ r ≠ r0 → (path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[r -- _type] .type, st)))
Proof step 5:
¬ r ≠ r0, validref(str, r, ty, st), path∃(str0, r, r0, st), level⊆(str, str0) ⊦ path∃(str, r, r0, st)
Proof step 17:
r ≠ r0, validref(str, r, ty, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str0, refs, st[r -- _type] .type, st), level⊆(str, str0) ⊦ path∃(str, r, r0, st)

Rule argument

[x0lemmaarg: spec-name: instance-name: lemma-name: pathex-def-nodups antecedent: succedent: r ≠ r0 → ( path∃(str, r, r0, st) ↔ (∃ refs. refs ≠ [] ∧ refs .first = r ∧ refs .last = r0 ∧ ¬ dups(refs) ∧ validrefs(str, refs, st[r -- _type] .type, st))) substlist: [st, r0, r, str] → [st, r0, r, str0] add preconditions?: F is rewrite lemma?: T current spec?: T rotate lemma?: F replace all?: T paths: ((0, 0, 1)) dummy: () ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r0, r, str0)]]