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Proof step 20 from proof hassharing-validrefs-dups (previous step)

∃ sks. fits((r ' + refs1 + r ') .first, sks, st) ∧ 1 + # sks = # (r ' + refs1 + r ') ∧ (∀ n. 0 < n ∧ n ≤ # sks → st[(r ' + refs1 + r ') .first -- firstn(n, sks)] .refval = (r ' + refs1 + r ')[n]), validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), path∃(str, r, r, st), ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
# sks = 1 + # refs1, validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), path∃(str, r, r, st), fits(r, sks, st), ∀ n. n ≠ 0 ∧ ¬ # sks < n → st[r -- firstn(n, sks)] .refval = (r ' + refs1 + r ')[n], ∀ st, str, refs2, r0. # (r0 ' + refs2) < # (r ' + refs1 + (r + refs)) → (validrefs(str, r0 ' + refs2, st[r0 -- _type] .type, st) → hasSharing(r0, st) ∨ ¬ r0 ∈ refs2 ∧ ¬ dups(refs2)) ⊦ hasSharing(r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m ≤ n ↔ n < m], [CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : ⊦ (r ' + refs) .first = r], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : n < n0 ⊦ m + ((* n)) = m0 + ((* n0)) ↔ m = m0 + (n0 - n)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0])]]