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Proof step 2 from proof from-append-add (
previous step
)
⊦ (σ to n) + σ0 from n1 + n = σ0 from n1, # σ0 < ⌜n1⌝ , # σ < ⌜n⌝
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
⊦ (σ to n) + σ0 from n1 + n = σ0 from n1, # σ < ⌜n⌝ , # σ0 < ⌜n1⌝
Rule argument
Additional information