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Proof step 5 from proof unchanged-junkref-level (previous step)

unchanged(str0, r, st, st1), level⊆(str, str0), junkrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ r ' + refs0 ≠ [] ∧ (r ' + refs0) .first = r ∧ junkrefs(str, r ' + refs0, st1[(r ' + refs0) .first -- _type] .type, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
unchanged(str0, r, st, st1), level⊆(str, str0), junkrefs(str, r ' + refs0, st[r -- _type] .type, st) ⊦ junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, c-08, ⊦ (r ' + refs) .first = r], [CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, e, ⊦ r ' ≠ []], [CSIMPNAMEDSEQ : list, references, app-02-ant, ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []])]]