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Proof step 25 from proof junkref-subset (previous step)

refs = r + refs1, junkrefs(str, refs, st1[r -- _type] .type, st1), st ⊆ st1 ⊦ refs = [], ∃ refs0. refs0 ≠ [] ∧ refs0 .first = r ∧ junkrefs(str, refs0, st[r -- _type] .type, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
st ⊆ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1) ⊦ ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]