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Proof step 37 from proof maxpath-array (previous step)

refs = r0 + refs1, st[r -- i '] = refval(r0), maxpath(str, r, st) = 1 + # refs0, maxpath(str, r0, st) = # refs, maxpath(str, r, st) ≤ # refs, 0 ≤ i, i < st[r -- _length] .intval, is_array(str, r, st), # refs0 < # refs, validrefs(str, r ' + refs0, st[r -- _type] .type, st), 0 < st[r -- _length] .intval, validrefs(str, refs, st[r0 -- _type] .type, st), is_valid_ref(str, r0, st) ⊦ refs = [], # validrefset(str, r0, st) = 0, validrefset(str, r0, st) = ∅, # validrefset(str, r, st) = 0, validrefset(str, r, st) = ∅, has_reffield(refs .last, st), dups(refs), junkref(str, r, st), cyclic(str, r, st), dups(refs0), r ∈ refs0, has_reffield((r ' + refs0) .last, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
st[r -- i '] = refval(r0), maxpath(str, r, st) = 1 + # refs0, maxpath(str, r0, st) = 1 + # refs1, is_valid_ref(str, r0, st), maxpath(str, r, st) ≤ maxpath(str, r0, st), 0 ≤ i, i < st[r -- _length] .intval, is_array(str, r, st), # refs0 ≤ # refs1, validrefs(str, r ' + refs0, st[r -- _type] .type, st), 0 < st[r -- _length] .intval, validrefs(str, r0 ' + refs1, st[r0 -- _type] .type, st) ⊦ # validrefset(str, r0, st) = 0, validrefset(str, r0, st) = ∅, # validrefset(str, r, st) = 0, validrefset(str, r, st) = ∅, junkref(str, r, st), cyclic(str, r, st), dups(refs0), r ∈ refs0, has_reffield((r ' + refs0) .last, st), has_reffield((r0 ' + refs1) .last, st), dups(refs1), r0 ∈ refs1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ # refsset ≠ 0 → refsset ≠ ∅], [CSIMPFORWARD : ⊦ refsset ≠ ∅ → # refsset ≠ 0], [CSIMPSEQ : ⊦ dups(r ' + refs) ↔ dups(refs) ∨ r ∈ refs], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]