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Proof step 12 from proof pathex-put-last (previous step)

r ' + refs0 = refs + r0 ' ∧ r ' + refs0 ≠ [], validrefs(str, r ' + refs0, st[r -- _type] .type, st[r0 -- sk, val]) ⊦ sk = _type, r = r0, path∃(str, r, r0, st), junkref(str, r, st), dups(refs0), r ∈ refs0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
r ' + refs0 = refs + refs0 .last ', r ∈ refs, validrefs(str, r ' + refs0, st[r -- _type] .type, st[refs0 .last -- sk, val]) ⊦ refs0 = [], sk = _type, r = refs0 .last, path∃(str, r, refs0 .last, st), junkref(str, r, st), dups(refs0), r ∈ refs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ r0 → (r ∈ r0 ' + refs ↔ r ∈ refs)], [CSIMPSEQ : ⊦ r ∈ refs → r ∈ refs + refs0], [CSIMPSEQ : ⊦ refs .last ∈ refs ↔ refs ≠ []], [CSIMPFORWARD : ⊦ refs + r ' = refs0 → r ∈ refs0], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPFORWARD : ⊦ r ' + refs = refs0 + r0 ' ∧ r ≠ r0 → r ∈ refs0 ∧ r0 ∈ refs], [CSIMPFORWARD : ⊦ r ' + refs = refs0 + r0 ' ∧ r ≠ r0 → refs .last = r0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []])]]