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Proof step 20 from proof junkref-subset (previous step)

r -- _type ∈ st → r -- _type ∈ st1 ∧ st[r -- _type] = st1[r -- _type], ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk], junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ junkrefs(str, r ' + refs1, st[r -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

User applied rule case distinction

And obtained the following new goals

Proof step 9:
∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk], junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ r -- _type ∈ st, junkrefs(str, r ' + refs1, st[r -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)
Proof step 19:
r -- _type ∈ st1 ∧ st[r -- _type] = st1[r -- _type], ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk], junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ junkrefs(str, r ' + refs1, st[r -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[CASEDARG : pos: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]