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Proof step 5 from proof okbytearray-is-byte (previous step)

¬ (okbytearray(st[r -- sks] .refval, st) ∧ 0 ≤ i ∧ i < st[st[r -- sks] .refval -- _length] .intval), 0 < st[r -- (sks + _length ')] .intval, 0 ≤ i, i < st[r -- (sks + _length ')] .intval, okbytearray(st[r -- sks] .refval, st) ⊦ sks = [], is_bytevalue(st[st[r -- sks] .refval -- i '])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
0 < st[r -- (sks + _length ')] .intval, 0 ≤ i, i < st[r -- (sks + _length ')] .intval, okbytearray(st[r -- sks] .refval, st), st[st[r -- sks] .refval -- _length] .intval ≤ i, st[st[r -- sks] .refval -- _length] .intval < st[r -- (sks + _length ')] .intval ⊦ sks = [], is_bytevalue(st[st[r -- sks] .refval -- i '])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ j < j0 → i < j0], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i ≤ j → (¬ ¬ j < i ↔ false)], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]