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Proof step 13 from proof hassharing-validrefs-dups (previous step)

st[r -- sks] .refval = r, # 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)) ⊦ sks = [], fits(r, sks, st) ∧ fits(r, sks + sks, st) ∧ sks ≠ sks + sks ∧ sks ≠ [] ∧ sks + sks ≠ [] ∧ st[r -- sks] .refval ≠ jvmref ∧ st[r -- sks] = st[r -- (sks + sks)]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# sks = 1 + # refs1, st[r -- sks] .refval = r, validrefs(str, r ' + refs1 + r ', st[r -- _type] .type, st), fits(r, sks, st), is_valid_ref(str, r, 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)) ⊦ sks = [], r = jvmref, st[r -- sks] = st[r -- (sks + sks)] ∧ fits(r, sks + sks, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ path∃(str, r, r, st) ↔ r = jvmref ∨ r ≠ jvmref ∧ is_valid_ref(str, r, st)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' ≠ []], [CSIMPSEQ : ⊦ ¬ refs + refs0 ≠ [] ↔ refs = [] ∧ refs0 = []], [CSIMPSEQ : ⊦ validrefs (str, jvmref ' + refs, ty, st) ↔ refs = []], [CSIMPSEQ : ⊦ ¬ sks + sks0 ≠ [] ↔ sks = [] ∧ sks0 = []], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks = sks + sks0 ↔ sks0 = []])]]