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Proof step 9 from proof junkref-subset (previous step)

∀ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1), ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk] ⊦ r -- _type ∈ st, junkrefs(str, r ' + refs1, noval .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, noval .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ ∅[rk] = noval], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]