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Proof step 17 from proof cont-monotone (previous step)

f(S0)(x), S0 ⊆ S ⊦ f(S)(x), S0 ⊆ (m = 0 ⊃ S0;S), (m = 0 ⊃ S0;S) ⊆ S0

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: subset-refl: ⊦ S ⊆ S; from specification monotone-set-fixpoint