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Proof step 28 from proof fits-put-other (previous step)

r -- sk0 ≠ r0 -- sk, ∀ r, st, r0, sk, val. (fits(r, sks0, st[r0 -- sk, val]) ↔ fits(r, sks0, st)) ∨ sk = _length ∨ sk = _type ∨ (∃ sks2, sks1. sks0 = sks2 + sk ' + sks1 ∧ st[r -- sks2] .refval = r0 ∧ (∀ sks3. sks3 + sk ' ⊑ sks2 → st[r -- sks3] .refval ≠ st[r -- sks2] .refval)), is_classtype(st[r -- _type] .type) → sk0 ∈ instfields(st[r -- _type] .type .class), r ≠ r0 ∨ sk0 ≠ sk ⊦ r = jvmref, sk = _length, sk = _type, is_referencevalue(st[r -- sk0]) ∧ (is_arraytype(st[r -- _type] .type) → is_indexkey(sk0) ∧ 0 ≤ sk0 .index ∧ sk0 .index < st[r -- _length] .intval) ∧ fits(st[r -- sk0] .refval, sks0, st[r0 -- sk, val]) ↔ is_referencevalue(st[r -- sk0]) ∧ (is_arraytype(st[r -- _type] .type) → ¬ is_fskey(sk0) ∧ ¬ sk0 .index < 0 ∧ ¬ st[r -- _length] .intval ≤ sk0 .index) ∧ fits(st[r -- sk0] .refval, sks0, st), ∃ sks, sks1. sk0 ' + sks0 = sks + sk ' + sks1 ∧ st[r -- sks] .refval = r0 ∧ (∀ sks3. sks3 + sk ' ⊑ sks → st[r -- sks3] .refval ≠ st[r -- sks] .refval)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPNAMEDSEQ : refkey, , inj-01, ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)])]]