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Proof step 18 from proof cont-monotone (
previous step
)
n ≠ 0, f(S0)(x), S0 ⊆ S ⊦ S ⊆ (m = 0 ⊃ S0;S), (m = 0 ⊃ S0;S) ⊆ (n = 0 ⊃ S0;S), f(S)(x)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 19
:
f(S0)(x), S0 ⊆ S ⊦ n = 0, f(S)(x), S ⊆ (m = 0 ⊃ S0;S), (m = 0 ⊃ S0;S) ⊆ S
Rule argument
Additional information