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Proof step 2 from proof map-length (previous step)

∀ intfun. # mapFun(intfun, ints0) = # ints0 ⊦ # mapFun(intfun, i + ints0) = #(i + ints0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ intfun. # mapFun(intfun, ints0) = # ints0 ⊦ # mapFun(intfun, ints0) = # ints0

Rule argument

Additional information

Used simplifier rules: mapFun-rec: ⊦ mapFun(intfun, i ' + ints) = intfun(i) ' + mapFun(intfun, ints);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%len-02: ⊦ # i ' = 0 +1; from specification list[intlista]%len: ⊦ #(ints + ints0) = # ints + # ints0; from specification list[intlista]%a: ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1; from specification list[intlista]%c: ⊦ i + ints = i ' + ints; from specification list[intlista]