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Proof step 19 from proof junkref-subset (previous step)

r -- _type ∈ st1 ∧ st[r -- _type] = st1[r -- _type], ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk], 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
st[r -- _type] = st1[r -- _type], r -- _type ∈ st1, junkrefs(str, r ' + refs1, st1[r -- _type] .type, st1), r ∈ st1, ∀ rk. rk ∈ st → rk ∈ st1 ∧ st[rk] = st1[rk] ⊦ junkrefs(str, r ' + refs1, st1[r -- _type] .type, st), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st[refs .first -- _type] .type, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1])]]