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Proof step 31 from proof RED-correct (previous step)

〈RED1(t; tn)〉 (tn = none ⊃ irred(t); t => tn .t), 〈if tn ≠ none then RED(tn .t; t0): kappa0 else t0 := t〉 t0 = t1, 〈RED1(t; tn0): kappa0 〉 tn0 = tn, ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), ¬ t =>* t1 ∨ ¬ irred(t1) ⊦

Heuristic split applied rule split left

And obtained the following new goals

Proof step 30:
〈RED1(t; tn)〉 tn = tn1, 〈if tn ≠ none then RED(tn .t; t0): kappa0 else t0 := t〉 t0 = t1, 〈RED1(t; tn0): kappa0 〉 tn0 = tn, ∀ t0, t2. ¬ 〈RED(t0; t'): kappa0 〉 t' = t2 ∨ t0 =>* t2 ∧ irred(t2), ¬ t =>* t1 ∨ ¬ irred(t1), tn1 = none ⊃ irred(t) ; t => tn1 .t ⊦

Rule argument

[fmapos-arg: pos: left 1]

Additional information

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