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Proof step 23 from proof junkref-subset (previous step)

st ⊆ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ r ' + refs1 ≠ [] ∧ (r ' + refs1) .first = r ∧ junkrefs(str, r ' + refs1, st[(r ' + refs1) .first -- _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 22:
st ⊆ st1, 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

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []])]]