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Proof step 12 from proof termination-RED (previous step)

∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), ¬ hasS(t), tn0 = none ⊦ 〈RED1(t; tn0)〉〈if tn0 ≠ none then RED(tn0 .t; t0) else t0 := t〉 true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ t0. #(t0) < #(t) → (¬ hasS(t0) → 〈RED(t0; t')〉 true), ¬ hasS(t) ⊦ 〈RED1(t; tn0)〉〈if tn0 ≠ none then RED(tn0 .t; t0) else t0 := t〉 true

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]