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Proof step 19 from proof fits-append (previous step)

sks1 ≠ [], ∀ r, sks0, st. fits(r, sks1 + sks0, st) ↔ fits(r, sks1, st) ∧ fits(st[r -- sks1] .refval, sks0, st) ⊦ fits(r, (sk ' + sks1) + sks0, st) ↔ fits(r, sk ' + sks1, st) ∧ fits(st[st[r -- sk] .refval -- sks1] .refval, sks0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
∀ r, sks0, st. fits(r, sks1 + sks0, st) ↔ fits(r, sks1, st) ∧ fits(st[r -- sks1] .refval, sks0, st) ⊦ sks1 = [], fits(r, (sk ' + sks1) + sks0, st) ↔ fits(r, sk ' + sks1, st) ∧ fits(st[st[r -- sk] .refval -- sks1] .refval, sks0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]