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Proof step 46 from proof fits-in (previous step)

∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ sk + sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sk + sks0, st) → r -- (sk + sks0) ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 45:
fits(r, sk ' + sks0, st), ∀ str, r, st. sks0 ≠ [] ∧ ¬ junkref(str, r, st) ∧ fits(r, sks0, st) → r -- sks0 ∈ st ⊦ junkref(str, r, st), r -- (sk ' + sks0) ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sk ' + sks ≠ []], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]