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Proof step 134 from proof unchanged-junkrefs-level (previous step)

∀ str0, st, st1, str, ty. unchanged(str0, refs0 .first, st, st1) ∧ junkrefs(str, refs0, ty, st) ∧ level⊆(str, str0) → junkrefs(str, refs0, ty, st1) ⊦ unchanged(str0, (r + refs0) .first, st, st1) ∧ junkrefs(str, r + refs0, ty, st) ∧ level⊆(str, str0) → junkrefs(str, r + refs0, ty, st1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 133:
level⊆(str, str0), unchanged(str0, r, st, st1), junkrefs(str, r ' + refs0, ty, st), ∀ str0, st, st1, str, ty. unchanged(str0, refs0 .first, st, st1) ∧ junkrefs(str, refs0, ty, st) ∧ level⊆(str, str0) → junkrefs(str, refs0, ty, st1) ⊦ junkrefs(str, r ' + refs0, ty, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : list, references, a, ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPNAMEDSEQ : list, references, c-08, ⊦ (r ' + refs) .first = r], [CSIMPNAMEDSEQ : list, references, c, ⊦ r + refs = r ' + refs])]]