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Proof step 57 from proof junkrefs-put-nopathex-equiv (previous step)

is_valid_ref(str, r0, st), ∀ str0, r2, st0, sk0, ty0, val0. ¬ path∃(str0, acc .first, r2, st0) ∧ is_valid_ref(str0, r2, st0) ∧ sk0 ≠ _type ∧ acc ≠ [] → (junkrefs(str0, acc, ty0, st0[r2 -- sk0, val0]) ↔ junkrefs(str0, acc, ty0, st0)) ⊦ sk = _type, junkrefs(str, r1 ' + acc, ty, st[r0 -- sk, val]) ↔ junkrefs(str, r1 ' + acc, ty, st), path∃(str, r1, r0, st)

User applied rule case distinction

And obtained the following new goals

Proof step 2:
junkrefs(str, r1 ' + acc, ty, st[r0 -- sk, val]), is_valid_ref(str, r0, st), ∀ str0, r2, st0, sk0, ty0, val0. ¬ path∃(str0, acc .first, r2, st0) ∧ is_valid_ref(str0, r2, st0) ∧ sk0 ≠ _type ∧ acc ≠ [] → (junkrefs(str0, acc, ty0, st0[r2 -- sk0, val0]) ↔ junkrefs(str0, acc, ty0, st0)) ⊦ junkrefs(str, r1 ' + acc, ty, st), sk = _type, path∃(str, r1, r0, st)
Proof step 56:
junkrefs(str, r1 ' + acc, ty, st), is_valid_ref(str, r0, st), ∀ str0, r2, st0, sk0, ty0, val0. ¬ path∃(str0, acc .first, r2, st0) ∧ is_valid_ref(str0, r2, st0) ∧ sk0 ≠ _type ∧ acc ≠ [] → (junkrefs(str0, acc, ty0, st0[r2 -- sk0, val0]) ↔ junkrefs(str0, acc, ty0, st0)) ⊦ junkrefs(str, r1 ' + acc, ty, st[r0 -- sk, val]), sk = _type, path∃(str, r1, r0, st)

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]